Microsoft Technologies: SQL Server Queries

TABLE SALESPEOPLE

SNUM SNAME CITY COMM

1001 Peel London .12

1002 Serres San Jose .13

1004 Motika London .11

1007 Rafkin Barcelona .15

1003 Axelrod New york .1

TABLE CUST

CNUM CNAME CITY RATING SNUM

2001 Hoffman London 100 1001

2002 Giovanne Rome 200 1003

2003 Liu San Jose 300 1002

2004 Grass Brelin 100 1002

2006 Clemens London 300 1007

2007 Pereira Rome 100 1004

ORDERS

ONUM AMT ODATE CNUM SNUM

3001 18.69 03-OCT-94 2008 1007

3003 767.19 03-OCT-94 2001 1001

3002 1900.10 03-OCT-94 2007 1004

3005 5160.45 03-OCT-94 2003 1002

3006 1098.16 04-OCT-94 2008 1007

3009 1713.23 04-OCT-94 2002 1003

3007 75.75 05-OCT-94 2004 1002

3008 4723.00 05-OCT-94 2006 1001

3010 1309.95 06-OCT-94 2004 1002

3011 9891.88 06-OCT-94 2006 1001

Problems :

1. Display snum,sname,city and comm of all salespeople.

Select snum, sname, city, comm

from salespeople;

2. Display all snum without duplicates from all orders.

Select distinct snum

from orders;

3. Display names and commissions of all salespeople in london.

Select sname,comm

from salespeople

where city = ‘London’;

4. All customers with rating of 100.

Select cname

from cust

where rating = 100;

5. Produce orderno, amount and date form all rows in the order table.

Select ordno, amt, odate

from orders;

6. All customers in San Jose, who have rating more than 200.

Select cname

from cust

where rating > 200;

7. All customers who were either located in San Jose or had a rating above 200.

Select cname

from cust

where city = ‘San Jose’ or

rating > 200;

8. All orders for more than $1000.

Select *

from orders

where amt > 1000;

9. Names and citires of all salespeople in london with commission above 0.10.

Select sname, city

from salepeople

where comm > 0.10 and

city = ‘London’;

10. All customers excluding those with rating <= 100 unless they are located in Rome.

Select cname

from cust

where rating <= 100 or

city = ‘Rome’;

11. All salespeople either in Barcelona or in london.

Select sname, city

from salespeople

where city in (‘Barcelona’,’London’);

12. All salespeople with commission between 0.10 and 0.12. (Boundary values should be excluded)

Select sname, comm

from salespeople

where comm > 0.10 and comm < 0.12;

13. All customers with NULL values in city column.

Select cname

from cust

where city is null;

14. All orders taken on Oct 3^Rdand Oct 4^th1994.

Select *

from orders

where odate in (‘03-OCT-94’,’04-OCT-94’);

15. All customers serviced by peel or Motika.

Select cname

from cust, orders

where orders.cnum = cust.cnum and

orders.snum in ( select snum

from salespeople

where sname in 'Peel','Motika'));

16. All customers whose names begin with a letter from A to B.

Select cname

from cust

where cname like ‘A%’ or

cname like ‘B%’;

17. All orders except those with 0 or NULL value in amt field.

Select onum

from orders

where amt != 0 or

amt is not null;

18. Count the number of salespeople currently listing orders in the order table.

Select count(distinct snum)

from orders;

19. Largest order taken by each salesperson, datewise.

Select odate, snum, max(amt)

from orders

group by odate, snum

order by odate,snum;

20. Largest order taken by each salesperson with order value more than $3000.

Select odate, snum, max(amt)

from orders

where amt > 3000

group by odate, snum

order by odate,snum;

21. Which day had the hightest total amount ordered.

Select odate, amt, snum, cnum

from orders

where amt = (select max(amt)

from orders)

22. Count all orders for Oct 3^rd.

Select count(*)

from orders

where odate = ‘03-OCT-94’;

23. Count the number of different non NULL city values in customers table.

Select count(distinct city)

from cust;

24. Select each customer’s smallest order.

Select cnum, min(amt)

from orders

group by cnum;

25. First customer in alphabetical order whose name begins with G.

Select min(cname)

from cust

where cname like ‘G%’;

26. Get the output like “ For dd/mm/yy there are ___ orders.

Select 'For ' || to_char(odate,'dd/mm/yy') || ' there are '||

count(*) || ' Orders'

from orders

group by odate;

27. Assume that each salesperson has a 12% commission. Produce order no., salesperson no., and amount of salesperson’s commission for that order.

Select onum, snum, amt, amt * 0.12

from orders

order by snum;

28. Find highest rating in each city. Put the output in this form. For the city (city), the highest rating is : (rating).

Select 'For the city (' || city || '), the highest rating is : (' ||

max(rating) || ')'

from cust

group by city;

29. Display the totals of orders for each day and place the results in descending order.

Select odate, count(onum)

from orders

group by odate

order by count(onum);

30. All combinations of salespeople and customers who shared a city. (ie same city).

Select sname, cname

from salespeople, cust

where salespeople.city = cust.city;

31. Name of all customers matched with the salespeople serving them.

Select cname, sname

from cust, salespeople

where cust.snum = salespeople.snum;

32. List each order number followed by the name of the customer who made the order.

Select onum, cname

from orders, cust

where orders.cnum = cust.cnum;

33. Names of salesperson and customer for each order after the order number.

Select onum, sname, cname

from orders, cust, salespeople

where orders.cnum = cust.cnum and

orders.snum = salespeople.snum;

34. Produce all customer serviced by salespeople with a commission above 12%.

Select cname, sname, comm

from cust, salespeople

where comm > 0.12 and

cust.snum = salespeople.snum;

35. Calculate the amount of the salesperson’s commission on each order with a rating above 100.

Select sname, amt * comm

from orders, cust, salespeople

where rating > 100 and

salespeople.snum = cust.snum and

salespeople.snum = orders.snum and

cust.cnum = orders.cnum

36. Find all pairs of customers having the same rating.

Select a.cname, b.cname,a.rating

from cust a, cust b

where a.rating = b.rating and

a.cnum != b.cnum

37. Find all pairs of customers having the same rating, each pair coming once only.

Select a.cname, b.cname,a.rating

from cust a, cust b

where a.rating = b.rating and

a.cnum != b.cnum and

a.cnum < b.cnum;

38. Policy is to assign three salesperson to each customers. Display all such combinations.

Select cname, sname

from salespeople, cust

where sname in ( select sname

from salespeople

where rownum <= 3)

order by cname;

39. Display all customers located in cities where salesman serres has customer.

Select cname

from cust

where city = ( select city

from cust, salespeople

where cust.snum = salespeople.snum and sname = 'Serres');

Select cname

from cust

where city in ( select city

from cust, orders

where cust.cnum = orders.cnum and

orders.snum in ( select snum

from salespeople

where sname = 'Serres'));

40. Find all pairs of customers served by single salesperson.

Select cname from cust

where snum in (select snum from cust

group by snum

having count(snum) > 1);

Select distinct a.cname

from cust a ,cust b

where a.snum = b.snum and a.rowid != b.rowid;

41. Produce all pairs of salespeople which are living in the same city. Exclude combinations of salespeople with themselves as well as duplicates with the order reversed.

Select a.sname, b.sname

from salespeople a, salespeople b

where a.snum > b.snum and

a.city = b.city;

42. Produce all pairs of orders by given customer, names that customers and eliminates duplicates.

Select c.cname, a.onum, b.onum

from orders a, orders b, cust c

where a.cnum = b.cnum and

a.onum > b.onum and

c.cnum = a.cnum;

43. Produce names and cities of all customers with the same rating as Hoffman.

Select cname, city

from cust

where rating = (select rating

from cust

where cname = 'Hoffman')

and cname != 'Hoffman';

44. Extract all the orders of Motika.

Select Onum

from orders

where snum = ( select snum

from salespeople

where sname = ‘Motika’);

45. All orders credited to the same salesperson who services Hoffman.

Select onum, sname, cname, amt

from orders a, salespeople b, cust c

where a.snum = b.snum and

a.cnum = c.cnum and

a.snum = ( select snum

from orders

where cnum = ( select cnum

from cust

where cname = 'Hoffman'));

46. All orders that are greater than the average for Oct 4.

Select *

from orders

where amt > ( select avg(amt)

from orders

where odate = '03-OCT-94');

47. Find average commission of salespeople in london.

Select avg(comm)

from salespeople

where city = ‘London’;

48. Find all orders attributed to salespeople servicing customers in london.

Select snum, cnum

from orders

where cnum in (select cnum

from cust

where city = 'London');

49. Extract commissions of all salespeople servicing customers in London.

Select comm

from salespeople

where snum in (select snum

from cust

where city = ‘London’);

50. Find all customers whose cnum is 1000 above the snum of serres.

Select cnum, cname from cust

where cnum > ( select snum+1000

from salespeople

where sname = 'Serres');

51. Count the customers with rating above San Jose’s average.

Select cnum, rating

from cust

where rating > ( select avg(rating)

from cust

where city = 'San Jose');

52. Obtain all orders for the customer named Cisnerous. (Assume you don’t know his customer no. (cnum)).

Select onum, odate

from orders

where cnum = ( select cnum

from cust

where cname = ‘Cisnerous’);

53. Produce the names and rating of all customers who have above average orders.

Select max(b.cname), max(b.rating), a.cnum

from orders a, cust b

where a.cnum = b.cnum

group by a.cnum

having count(a.cnum) > ( select avg(count(cnum))

from orders

group by cnum);

54. Find total amount in orders for each salesperson for whom this total is greater than the amount of the largest order in the table.

Select snum,sum(amt)

from orders

group by snum

having sum(amt) > ( select max(amt)

from orders);

55. Find all customers with order on 3^rd Oct.

Select cname

from cust a, orders b

where a.cnum = b.cnum and

odate = ‘03-OCT-94’;

56. Find names and numbers of all salesperson who have more than one customer.

Select sname, snum

from salespeople

where snum in ( select snum

from cust

group by snum

having count(snum) > 1 );

57. Check if the correct salesperson was credited with each sale.

Select onum, a.cnum, a.snum, b.snum

from orders a, cust b

where a.cnum = b.cnum and

a.snum != b.snum;

58. Find all orders with above average amounts for their customers.

select onum, cnum, amt

from orders a

where amt > ( select avg(amt)

from orders b

where a.cnum = b.cnum

group by cnum);

59. Find the sums of the amounts from order table grouped by date, eliminating all those dates where the sum was not at least 2000 above the maximum amount.

Select odate, sum(amt)

from orders a

group by odate

having sum(amt) > ( select max(amt)

from orders b

where a.odate = b.odate

group by odate);

60. Find names and numbers of all customers with ratings equal to the maximum for their city.

Select a.cnum, a.cname

from cust a

where a.rating = ( select max(rating)

from cust b

where a.city = b.city);

61. Find all salespeople who have customers in their cities who they don’t service. ( Both way using Join and Correlated subquery.)

Select distinct cname

from cust a, salespeople b

where a.city = b.city and

a.snum != b.snum;

Select cname

from cust

where cname in ( select cname

from cust a, salespeople b

where a.city = b.city and

a.snum != b.snum );

62. Extract cnum,cname and city from customer table if and only if one or more of the customers in the table are located in San Jose.

Select * from cust

where 2 < (select count(*)

from cust

where city = 'San Jose');

63. Find salespeople no. who have multiple customers.

Select snum

from cust

group by snum

having count(*) > 1;

64. Find salespeople number, name and city who have multiple customers.

Select snum, sname, city

from salespeople

where snum in ( select snum

from cust

group by snum

having count(*) > 1);

65. Find salespeople who serve only one customer.

Select snum

from cust

group by snum

having count(*) = 1;

66. Extract rows of all salespeople with more than one current order.

Select snum, count(snum)

from orders

group by snum

having count(snum) > 1;

67. Find all salespeople who have customers with a rating of 300. (use EXISTS)

Select a.snum

from salespeople a

where exists ( select b.snum

from cust b

where b.rating = 300 and

a.snum = b.snum)

68. Find all salespeople who have customers with a rating of 300. (use Join).

Select a.snum

from salespeople a, cust b

where b.rating = 300 and

a.snum = b.snum;

69. Select all salespeople with customers located in their cities who are not assigned to them. (use EXISTS).

Select snum, sname

from salespeople

where exists ( select cnum

from cust

where salespeople.city = cust.city and

salespeople.snum != cust.snum);

70. Extract from customers table every customer assigned the a salesperson who currently has at least one other customer ( besides the customer being selected) with orders in order table.

Select a.cnum, max(c.cname)

from orders a, cust c

where a.cnum = c.cnum

group by a.cnum,a.snum

having count(*) < ( select count(*)

from orders b

where a.snum = b.snum)

order by a.cnum;

71. Find salespeople with customers located in their cities ( using both ANY and IN).

Select sname

from salespeople

where snum in ( select snum from cust

where salespeople.city = cust.city and

salespeople.snum = cust.snum);

Select sname

from salespeople

where snum = any ( select snum

from cust

where salespeople.city = cust.city and

salespeople.snum = cust.snum);

72. Find all salespeople for whom there are customers that follow them in alphabetical order. (Using ANY and EXISTS)

Select sname

from salespeople

where sname < any ( select cname

from cust

where salespeople.snum = cust.snum);

Select sname

from salespeople

where exists ( select cname

from cust

where salespeople.snum = cust.snum and

salespeople.sname < cust.cname);

73. Select customers who have a greater rating than any customer in rome.

Select a.cname

from cust a

where city = 'Rome' and

rating > ( select max(rating)

from cust

where city != 'Rome');

74. Select all orders that had amounts that were greater that atleast one of the orders from Oct 6^th.

Select onum, amt

from orders

where odate != '06-oct-94' and

amt > ( select min(amt)

from orders

where odate = '06-oct-94');

75. Find all orders with amounts smaller than any amount for a customer in San Jose. (Both using ANY and without ANY)

Select onum, amt

from orders

where amt < any ( select amt

from orders, cust

where city = 'San Jose' and

orders.cnum = cust.cnum);

Select onum, amt

from orders

where amt < ( select max(amt)

from orders, cust

where city = 'San Jose' and

orders.cnum = cust.cnum);

76. Select those customers whose ratings are higher than every customer in Paris. ( Using both ALL and NOT EXISTS).

Select * from cust

where rating > any (select rating from cust

where city = 'Paris');

Select *

from cust a

where not exists ( select b.rating from cust b

where b.city != 'Paris' and

b.rating > a.rating);

77. Select all customers whose ratings are equal to or greater than ANY of the Seeres.

Select cname, sname

from cust, salespeople

where rating >= any ( select rating

from cust

where snum = (select snum

from salespeople

where sname = 'Serres'))

and sname != 'Serres'

and salespeople.snum(+) = cust.snum;

78. Find all salespeople who have no customers located in their city. ( Both using ANY and ALL)

Select sname

from salespeople

where snum in ( select snum

from cust

where salespeople.city != cust.city and

salespeople.snum = cust.snum);

Select sname

from salespeople

where snum = any ( select snum

from cust

where salespeople.city != cust.city and

salespeople.snum = cust.snum);

79. Find all orders for amounts greater than any for the customers in London.

Select onum, amt

from orders

where amt > any ( select amt

from orders, cust

where city = ‘London’ and

orders.cnum = cust.cnum);

80. Find all salespeople and customers located in london.

Select sname, cname

from cust, salespeople

where cust.city = 'London' and

salespeople.city = 'London' and

cust.snum = salespeople.snum;

81. For every salesperson, dates on which highest and lowest orders were brought.

Select a.amt, a.odate, b.amt, b.odate

from orders a, orders b

where (a.amt, b.amt) in (select max(amt), min(amt)

from orders

group by snum);

82. List all of the salespeople and indicate those who don’t have customers in their cities as well as those who do have.

Select snum, city, 'Customer Present'

from salespeople a

where exists ( select snum from cust

where a.snum = cust.snum and

a.city = cust.city)

UNION

select snum, city, 'Customer Not Present'

from salespeople a

where exists ( select snum from cust c

where a.snum = c.snum and

a.city != c.city and

c.snum not in ( select snum

from cust

where a.snum = cust.snum and

a.city = cust.city));

83. Append strings to the selected fields, indicating weather or not a given salesperson was matched to a customer in his city.

Select a.cname, decode(a.city,b.city,'Matched','Not Matched')

from cust a, salespeople b

where a.snum = b.snum;

84. Create a union of two queries that shows the names, cities and ratings of all customers. Those with a rating of 200 or greater will also have the words ‘High Rating’, while the others will have the words ‘Low Rating’.

Select cname, cities, rating, ‘Higher Rating’

from cust

where rating >= 200

UNION

Select cname, cities, rating, ‘Lower Rating’

from cust

where rating < 200;

85. Write command that produces the name and number of each salesperson and each customer with more than one current order. Put the result in alphabetical order.

Select 'Customer Number ' || cnum "Code ",count(*)

from orders

group by cnum

having count(*) > 1

UNION

select 'Salesperson Number '||snum,count(*)

from orders

group by snum

having count(*) > 1;

86. Form a union of three queries. Have the first select the snums of all salespeople in San Jose, then second the cnums of all customers in San Jose and the third the onums of all orders on Oct. 3. Retain duplicates between the last two queries, but eliminates and redundancies between either of them and the first.

Select 'Customer Number ' || cnum "Code "

from cust

where city = 'San Jose'

UNION

select 'Salesperson Number '||snum

from salespeople

where city = 'San Jose'

UNION ALL

select 'Order Number '|| onum

from Orders

where odate = '03-OCT-94';

87. Produce all the salesperson in London who had at least one customer there.

Select snum, sname

from salespeople

where snum in ( select snum

from cust

where cust.snum = salespeople.snum and

cust.city = 'London')

and city = ‘London’;

88. Produce all the salesperson in London who did not have customers there.

Select snum, sname

from salespeople

where snum in ( select snum

from cust

where cust.snum = salespeople.snum and

cust.ci

89. We want to see salespeople matched to their customers without excluding those salesperson who were not currently assigned to any customers. (User OUTER join and UNION)

Select sname, cname

from cust, salespeople

where cust.snum(+) = salespeople.snum;

Select sname, cname

from cust, salespeople

where cust.snum = salespeople.snum

UNION

select distinct sname, 'No Customer'

from cust, salespeople

where 0 = (select count(*) from cust

where cust.snum = salespeople.snum);

##################################################################################################################################################################

SQL-QUERIES
1.
Display all the information of the EMP table?
A) select * from emp;
2.
Display unique Jobs from EMP table?
A)
select distinct job from emp;
B)
select unique job from emp;
3.
List the emps in the asc order of their Salaries?
A) select * from emp order by sal asc;
4.
List the details of the emps in asc order of the Dptnos and desc of Jobs?
A)select * from emp order by deptno asc,job desc;
5.
Display all the unique job groups in the descending order?
A)select distinct job from emp order by job desc;
6.
Display all the details of all ‘Mgrs’
A)Select * from emp where empno in ( select mgr from emp) ;
7.
List the emps who joined before 1981.
A) select * from emp where hiredate < (’01-jan-81’);
8.
List the Empno, Ename, Sal, Daily sal of all emps in the asc order of Annsal.
A) select empno ,ename ,sal,sal/30,12*sal annsal from emp order by annsal asc;
9.
Display the Empno, Ename, job, Hiredate, Exp of all Mgrs
A) select empno,ename ,job,hiredate, months_between(sysdate,hiredate) exp from emp where empno in (select mgr from emp);
10.
List the Empno, Ename, Sal, Exp of all emps working for Mgr 7369.
A) select empno,ename,sal,exp from emp where mgr = 7369;
11.
Display all the details of the emps whose Comm. Is more than their Sal.
A) select * from emp where comm. > sal;
12.
List the emps in the asc order of Designations of those joined after the second half of 1981.
A) select * from emp where hiredate > (’30-jun-81’) and to_char(hiredate,’YYYY’) = 1981 order by job asc;
13.
List the emps along with their Exp and Daily Sal is more than Rs.100.
A) select * from emp where (sal/30) >100;
14.
List the emps who are either ‘CLERK’ or ‘ANALYST’ in the Desc order.
A) select * from emp where job = ‘CLERK’ or job = ‘ANALYST’ order by job desc;
15.
List the emps who joined on 1-MAY-81,3-DEC-81,17-DEC-81,19-JAN-80 in asc order of seniority.
A) select * from emp where hiredate in (’01-may-81’,’03-dec-81’,’17-dec-81’,’19-jan-80’) order by hiredate asc;
16.
List the emp who are working for the Deptno 10 or20.
A) select * from emp where deptno = 10 or deptno = 20 ;
17.
List the emps who are joined in the year 81.
A) select * from emp where hiredate between ’01-jan-81’ and ’31-dec-81’;
18.
List the emps who are joined in the month of Aug 1980.
A)
select * from emp where hiredate between ’01-aug-80’ and ’31-aug-80’; (OR)
select * from emp where to_char(hiredate,’mon-yyyy’) =’aug-1980;
19.
List the emps Who Annual sal ranging from 22000 and 45000.
A) select * from emp where 12*sal between 22000 and 45000;
20.
List the Enames those are having five characters in their Names.
A) select ename from emp where length (ename) = 5;
21.
List the Enames those are starting with ‘S’ and with five characters.
A) select ename from emp where ename like ‘S%’ and length (ename) = 5;
22.
List the emps those are having four chars and third character must be ‘r’.
A) select * from emp where length(ename) = 4 and ename like ‘__R%’;
23.
List the Five character names starting with ‘S’ and ending with ‘H’.
A) select * from emp where length(ename) = 5 and ename like ‘S%H’;
24.
List the emps who joined in January.
A) select * from emp where to_char (hiredate,’mon’) = ‘jan’;
25.
List the emps who joined in the month of which second character is ‘a’.
A)
select * from emp where to_char(hiredate,’mon’) like ‘_a_’; (OR)
B) select * from emp where to_char(hiredate,’mon’) like ‘_a%’;
26.
List the emps whose Sal is four digit number ending with Zero.
A) select * from emp where length (sal) = 4 and sal like ‘%0’;
27.
List the emps whose names having a character set ‘ll’ together.
A) select * from emp where ename like ‘%LL%’;
28.
List the emps those who joined in 80’s.
A) select * from emp where to_char(hiredate,’yy’) like ‘8%’;
29.
List the emps who does not belong to Deptno 20.
A) select * from emp where deptno not in (20); (OR)
B) select * from emp where deptno != 20; (OR)
C) select * from emp where deptno <>20; (OR)
D) select * from emp where deptno not like ‘20’;
30.
List all the emps except ‘PRESIDENT’ & ‘MGR” in asc order of Salaries.
A)
Select * from emp where job not in (‘PRESIDENT’,’MANAGER’) order by sal asc;
B)
select * from emp where job not like ‘PRESIDENT’ and job not like ‘MANAGER’ order by sal asc;
C) Select * from emp where job != ‘PRESIDENT’ and job <> ‘MANAGER’ order by sal asc;
31.
List all the emps who joined before or after 1981.
A)
select * from emp where to_char (hiredate,’YYYY’) not in (‘1981’); (OR)
B)
select * from emp where to_char ( hiredate,’YYYY’) != ‘1981’; (OR)
C)
select * from emp where to_char(hiredate,’YYYY’) <> ‘1981’ ; (OR)
D) select * from emp where to_char (hiredate ,’YYYY’) not like ‘1981’;
32.
List the emps whose Empno not starting with digit78.
A) select * from emp where empno not like ‘78%’;
33.
List the emps who are working under ‘MGR’.
A) select e.ename || ‘ works for ‘ || m.ename from emp e ,emp m where e.mgr = m.empno ; (OR)
B) select e.ename || ‘ has an employee ‘|| m.ename from emp e , emp m where e.empno = m.mgr;
34.
List the emps who joined in any year but not belongs to the month of March.
A)
select * from emp where to_char (hiredate,’MON’) not in (‘MAR’); (OR)
B)
select * from emp where to_char (hiredate,’MON’) != ‘MAR’; (OR)
C)
select * from emp where to_char(hiredate,’MONTH’) not like ‘MAR%’ ; (OR)
D)
select * from emp where to_char(hiredate,’MON’) <> ‘MAR’;
35.
List all the Clerks of Deptno 20.
A)select * from emp where job =‘CLERK’ and deptno = 20;
36.
List the emps of Deptno 30 or 10 joined in the year 1981.
A) select * from emp where to_char(hiredate,’YYYY’) = ‘1981’ and (deptno =30 or deptno =10) ; (OR) select * from emp where to_char (hiredate,’YYYY’) in (‘1981’) and (deptno = 30 or deptno =10 ) ;
37.
Display the details of SMITH.
A) select * from emp where ename = ‘SMITH’ ;
38.
Display the location of SMITH.
A) select loc from emp e , dept d where e.ename = ‘SMITH’ and e.deptno = d.deptno ;
39.
List the total information of EMP table along with DNAME and Loc of all the emps Working Under ‘ACCOUNTING’ & ‘RESEARCH’ in the asc Deptno.
A)
select * from emp e ,dept d where (dname = ‘ACCOUNTING’ or dname =’RESEARCH’ ) and e.deptno = d.deptno order by e.deptno asc; (OR)
B)
select * from emp e ,dept d where d.dname in (‘ACCOUNTING’,’RESEARCH’) and e.deptno = d.deptno order by e.deptno asc;
40.
List the Empno, Ename, Sal, Dname of all the ‘MGRS’ and ‘ANALYST’ working in New York, Dallas with an exp more than 7 years without receiving the Comm asc order of Loc.
A)
select e.empno,e.ename,e.sal,d.dname from emp e ,dept d where d.loc in (‘NEW YORK’,’DALLAS’) and e.deptno = d.deptno and e.empno in (select e.empno from emp e where e.job in (‘MANAGER’,’ANALYST’) and (months_between(sysdate,e.hiredate)/12)> 7 and e.comm. is null)
order by d.loc asc;
41.
Display the Empno, Ename, Sal, Dname, Loc, Deptno, Job of all emps working at CJICAGO or working for ACCOUNTING dept with Ann Sal>28000, but the Sal should not be=3000 or 2800 who doesn’t belongs to the Mgr and whose no is having a digit ‘7’ or ‘8’ in 3rd position in the asc order of Deptno and desc order of job.
A) select E.empno,E.ename,E.sal,D.dname,D.loc,E.deptno,E.job
from emp E,dept D
where (D.loc = 'CHICAGO' or D.dname = 'ACCOUNTING') and E.deptno=D.deptno and E.empno in
(select E.empno from emp E where (12*E.sal) > 28000 and E.sal not in (3000,2800) and E.job !='MANAGER'
and ( E.empno like '__7%' or E.empno like '__8%'))
order by E.deptno asc , E.job desc;
42.
Display the total information of the emps along with Grades in the asc order.
A) select * from emp e ,salgrade s where e.sal between s.losal and s.hisal order by grade asc; (OR)
B) select * from emp e ,salgrade s where e.sal >= s.losal and e.sal <= s.hisal order by s.grade asc; (using between and is a bit simple)
43.
List all the Grade2 and Grade 3 emps.
A)
select * from emp e where e.empno in (select e.empno from emp e ,salgrade s where e.sal between s.losal and s.hisal and s.grade in(2,3)); (OR)
B) select * from emp e ,salgrade s where e.sal between s.losal and s.hisal and
s.grade in (2,3) ;
44.
Display all Grade 4,5 Analyst and Mgr.
A) select * from emp e, salgrade s where e.sal between s.losal and s.hisal and s.grade in (4,5) and e.empno in (select e.empno from emp e where e.job in (‘MANAGER’,’ANALYST’) );
45.
List the Empno, Ename, Sal, Dname, Grade, Exp, and Ann Sal of emps working for Dept10 or20.
A)
selectE.empno,E.ename,E.sal,S.grade,D.dname,(months_between(sysdate,E.hiredate)/12) "EXP" ,12*E.sal “ANN SAL”
from emp E,dept D ,salgrade S
where E.deptno in (10,20) and E.deptno = D.deptno and E.sal between S.losal and S.hisal ;
46.
List all the information of emp with Loc and the Grade of all the emps belong to the Grade range from 2 to 4 working at the Dept those are not starting with char set ‘OP’ and not ending with ‘S’ with the designation having a char ‘a’ any where joined in the year 1981 but not in the month of Mar or Sep and Sal not end with ‘00’ in the asc order of Grades
A) select e.empno,e.ename,d.loc,s.grade,e.sal from emp e ,dept d,salgrade s where e.deptno = d.deptno
and (d.dname not like 'OP%' and d.dname not like '%S') and e.sal between s.losal and s.hisal and s.grade in (2,3,4)
and empno in (select empno from emp where job like '%A%'and sal not like '' and (to_char (hiredate,'YYYY') = '1981'
and to_char(hiredate,'MON') not in ('MAR','SEP')));
47.
List the details of the Depts along with Empno, Ename or without the emps
A) select * from emp e,dept d where e.deptno(+)= d.deptno;
48.
List the details of the emps whose Salaries more than the employee BLAKE.
A) select * from emp where sal > (select sal from emp where ename = ‘BLAKE’);
49.
List the emps whose Jobs are same as ALLEN.
A) select * from emp where job = (select job from emp where ename = ‘ALLEN’);
50.
List the emps who are senior to King.
A)
select * from emp where hiredate < ( select hiredate from emp where ename = ‘KING’);
51.
List the Emps who are senior to their own MGRS.
A)
select * from emp w,emp m where w.mgr = m.empno and w.hiredate < m.hiredate ; (OR)
B)
select * from emp w,emp m where w.empno= m.mgr and
w.hiredate> m.hiredate;
52.
List the Emps of Deptno 20 whose Jobs are same as Deptno10.
A) select * from emp e ,dept d where d.deptno = 20 and e.deptno = d.deptno and e.job in ( select e.job from emp e,dept d where e.deptno = d.deptno and d.deptno =10);
53.
List the Emps whose Sal is same as FORD or SMITH in desc order of Sal.
A)
Select * from emp where sal in (select sal from emp where ( ename = ‘SMITH’ or ename = ‘FORD’ )) order by sal desc;
54.
List the emps Whose Jobs are same as MILLER or Sal is more than ALLEN.
A) select * from emp where job = (select job from emp where ename = ‘MILLER’ ) or sal>(select sal from emp where ename = ‘ALLEN’);
55.
List the Emps whose Sal is > the total remuneration of the SALESMAN.
A) select * from emp where sal >(select sum(nvl2(comm,sal+comm,sal)) from emp where job = ‘SALESMAN’);
56.
List the emps who are senior to BLAKE working at CHICAGO & BOSTON.
A)
select * from emp e ,dept d where d.loc in (‘CHICAGO’,’BOSTON’) and e.deptno = d.deptno and e.hiredate <(select e.hiredate from emp e where e.ename = ‘BLAKE’) ;
57.
List the Emps of Grade 3,4 belongs to the dept ACCOUNTING and RESEARCH whose Sal is more than ALLEN and exp more than SMITH in the asc order of EXP.
A)
select * from emp e where e.deptno in (select d.deptno from dept d where d.dname in (‘ACCOUNTING’,’RESEARCH’) ) and
e.sal >(select sal from emp where ename = ‘ALLEN’) and
e.hiredate <( select hiredate from emp where ename = ‘SMITH’) and
e.empno in (select e.empno from emp e ,salgrade s where e.sal between s.losal and s.hisal and s.grade in (3,4) )
order by e.hiredate desc;
58.
List the emps whose jobs same as SMITH or ALLEN.
A)
select * from emp where job in (select job from emp where ename = ‘SMITH’ or ename = ‘ALLEN’); (OR)
B) select * from emp where job in (select job from emp where ename in (‘SMITH’,’ALLEN’);
59.
Write a Query to display the details of emps whose Sal is same as of
a)
Employee Sal of EMP1 table.
b)
¾ Sal of any Mgr of EMP2 table.
c)
The sal of any person with exp of 5 years belongs to the sales dept of emp3 table.
d)
Any grade 2 employee of emp4 table.
e)
Any grade 2 and 3 employee working fro sales dept or operations dept joined in 89.
60.
Any jobs of deptno 10 those that are not found in deptno 20.
A) select e.job from emp e where e.deptno = 10 and e.job not in (select job from emp where deptno =20);
61.
List of emps of emp1 who are not found in emp2.
62.
Find the highest sal of EMP table.
A) select max(sal) from emp;
63.
Find details of highest paid employee.
A)
select * from emp where sal in (select max(sal) from emp);
64.
Find the highest paid employee of sales department.
A) select * from emp where sal in (select max(sal) from emp where deptno in (select d.deptno from
dept d where d.dname = 'SALES'));
65.
List the most recently hired emp of grade3 belongs to location CHICAGO.
A) select * from emp e where e.deptno in ( select d.deptno from dept d where d.loc = 'CHICAGO') and
e.hiredate in (select max(hiredate) from emp where empno in (select empno from emp e,salgrade s
where e.sal between s.losal and s.hisal and s.grade = 3)) ; (or)
select * from emp e,dept d where d.loc='chicago'
and hiredate in(select max(hiredate) from emp e,salgrade s
where sal between losal and hisal and grade=3);
66.
List the employees who are senior to most recently hired employee working under king.
A) select * from emp where hiredate < (select max(hiredate) from emp where mgr in
(select empno from emp where ename = 'KING')) ;
67.
List the details of the employee belongs to newyork with grade 3 to 5 except ‘PRESIDENT’ whose sal> the highest paid employee of Chicago in a group where there is manager and salesman not working under king
A) select * from emp where deptno in (select deptno from dept where dept.loc ='NEW YORK')
and empno in (select empno from emp e,salgrade s where e.sal between s.losal and s.hisal and
s.grade in (3,4,5) ) and job != 'PRESIDENT' and sal >(select max(sal) from emp where deptno in
(select deptno from dept where dept.loc = 'CHICAGO') and job in ('MANAGER','SALESMAN') and
mgr not in (select empno from emp where ename = 'KING'));
68.
List the details of the senior employee belongs to 1981.
A)
select * from emp where hiredate in (select min(hiredate) from emp where to_char( hiredate,’YYYY’) = ‘1981’); (OR)
B)
select * from emp where hiredate = (select min(hiredate) from emp where to_char(hiredate,’YYYY’) = ‘1981’);
69.
List the employees who joined in 1981 with the job same as the most senior person of the year 1981.
A)select * from emp where job in (select job from emp where hiredate in
(select min(hiredate) from emp where to_char(hiredate,’YYYY’) =’1981’));
70.
List the most senior empl working under the king and grade is more than 3.
A) select * from emp where hiredate in (select min(hiredate) from emp where empno in
(select empno from emp e ,salgrade s where e.sal between s.losal and s.hisal and s.grade in (4,5)))
and mgr in (select empno from emp where ename = 'KING');
71.
Find the total sal given to the MGR.
A)
select sum (sal) from emp where job = ‘MANAGER’; (OR)
B) select sum(sal) from emp where empno in(select mgr from emp);
72.
Find the total annual sal to distribute job wise in the year 81.
A) select job,sum(12*sal) from emp where to_char(hiredate,'YYYY') = '1981'
group by job ;
73.
Display total sal employee belonging to grade 3.
A)
select sum(sal) from emp where empno
in (select empno from emp e ,salgrade s
where e.sal between s.losal and s.hisal and s.grade = 3)
74.
Display the average salaries of all the clerks.
A) select avg(sal) from emp where job = ‘CLERK’;
75.
List the employeein dept 20 whose sal is >the average sal 0f dept 10 emps.
A) select * from emp where deptno =20 and sal >(select avg (sal) from emp where deptno = 10);
76.
Display the number of employee for each job group deptno wise.
A)
select deptno ,job ,count(*) from emp group by deptno,job; (or)
B) select d.deptno,e.job,count(e.job) from emp e,dept d where e.deptno(+)=d.deptno group by e.job,d.deptno;
77.
List the manage rno and the number of employees working for those mgrs in the ascending Mgrno.
A)
select w.mgr ,count(*) from emp w,emp m
where w.mgr = m.empno
group by w.mgr
order by w.mgr asc;
78.
List the department,details where at least two emps are working
A)
select deptno ,count(*) from emp group by deptno
having count(*) >= 2;
79.
Display the Grade, Number of emps, and max sal of each grade.
A) select s.grade ,count(*),max(sal) from emp e,salgrade s where e.sal between s.losal and s.hisal
group by s.grade;
80.
Display dname, grade, No. of emps where at least two emps are clerks.
A) select d.dname,s.grade,count(*) from emp e,dept d,salgrade s where e.deptno = d.deptno and
e.job = 'CLERK' and e.sal between s.losal and s.hisal group by d.dname,s.grade having count(*) >= 2;
81.
List the details of the department where maximum number of emps are working.
A)
select * from dept where deptno in
(select deptno from emp group by deptno
having count(*) in
(select max(count(*)) from emp group by deptno) ); (OR)
B)
select d.deptno,d.dname,d.loc,count(*) from emp e ,dept d
where e.deptno = d.deptno group by d.deptno,d.dname,d..loc
having count(*) = (select max(count(*) ) from emp group by deptno);
82.
Display the emps whose manager name is jones.
A)
select * from emp where mgr in
(select empno from emp where ename = ‘JONES’); (OR)
B)
select * from emp where mgr =
(select empno from emp where ename = ‘JONES’);
83.
List the employees whose salary is more than 3000 after giving 20% increment.
A)
SELECT * FROM EMP WHERE (1.2*SAL) > 3000 ;
84.
List the emps with dept names.
A) select e.empno,e.ename,e.job,e.mgr,e.hiredate,e.sal,e.comm,e.deptno,d.dname
from emp e ,dept d where e.deptno = d.deptno;
85.
List the emps who are not working in sales dept.
A)
select * from emp where deptno not in
(select deptno from emp where dname = ‘SALES’);
86.
List the emps name ,dept, sal and comm. For those whose salary is between 2000 and 5000 while loc is Chicago.
A) select e.ename,e.deptno,e.sal,e.comm from emp e ,dept d where e.deptno = d.deptno and
d.loc = 'CHICAGO' and e.sal between 2000 and 5000;
87.
List the emps whose sal is greater than his managers salary
A) select * from emp w,emp m where w.mgr = m.empno and w.sal > m.sal;
88.
List the grade, EMP name for the deptno 10 or deptno 30 but sal grade is not 4 while they joined the company before ’31-dec-82’.
A) select s.grade ,e.ename from emp e,salgrade s where e.deptno in (10,20) and
hiredate < ('31-DEC-82') and (e.sal between s.losal and s.hisal and s.grade not in (4));
89.
List the name ,job, dname, location for those who are working as MGRS.
A)
select e.ename,e.job,d.dname,d.loc from emp e ,dept d
where e.deptno = d.deptno and
e.empno in (select mgr from emp ) ;
90.
List the emps whose mgr name is jones and also list their manager name.
A) select w.empno,w.ename,w.job,w.mgr,w.hiredate,w.sal,w.deptno,m.ename from emp w ,emp m
where w.mgr = m.empno and m.ename = 'JONES';
91.
List the name and salary of ford if his salary is equal to hisal of his grade.
A) select e.ename,e.sal from emp e ,salgrade s where e.ename = 'FORD' and e.sal between s.losal and s.hisal and e.sal = s.hisal ;
92.
Lit the name, job, dname ,sal, grade dept wise
A)
select e.ename,e.job,d.dname,e.sal,s.grade from emp e,dept d,salgrade s
where e.deptno = d.deptno and e.sal between s.losal and s.hisal
order by e.deptno ;
93.
List the emp name, job, sal, grade and dname except clerks and sort on the basis of highest sal.
A)
select e.ename,e.job,e.sal,s.grade,d.dname from emp e ,dept d ,salgrade s where e.deptno = d.deptno and e.sal between s.losal and s.hisal and
e.job not in('CLERK')
order by e.sal desc;
94.
List the emps name, job who are with out manager.
A) select e.ename,e.job from emp e where mgr is null;
95.
List the names of the emps who are getting the highest sal dept wise.
A)
select e.ename,e.deptno from emp e where e.sal in
(select max(sal) from emp group by deptno) ;
96.
List the emps whose sal is equal to the average of max and minimum
A) select * from emp where sal =(select (max(sal)+min(sal))/2 from emp);
97.
List the no. of emps in each department where the no. is more than 3.
A) select deptno,count(*) from emp group by deptno having count(*) < 3;
98.
List the names of depts. Where atleast 3 are working in that department.
A)
select d.dname,count(*) from emp e ,dept d where e.deptno = d.deptno
group by d.dname
having count(*) >= 3 ;
99.
List the managers whose sal is more than his employess avg salary.
A)
select * from emp m where m.empno in (select mgr from emp)
and m.sal > (select avg(e.sal) from emp e wheree.mgr = m.empno )
The subquery does the same as (
100.
List the name,salary,comm. For those employees whose net pay is greater than or equal to any other employee salary of the company.
A)
select e.ename,e.sal,e.comm from emp e where nvl2(e.comm.,e.sal+e.comm.,e.sal) >= any (select sal from emp); (OR)
B)
select ename,sal,comm. from emp where sal+nvl(comm.,0) >= any (select sal from emp);/
101.
List the emp whose sal<his manager but more than any other manager.
a) select distinct W.empno,W.ename,W.sal
from (select w.empno,w.ename,w.sal from emp w,emp m where
w.mgr = m.empno and w.sal<m.sal) W,
(select * from emp where empno in (select mgr from emp)) A
where W.sal > A.sal; (OR)
B) select * from emp w,emp m where w.mgr = m.empno and w.sal < m.sal
and w.sal > any (select sal from emp where empno in (select mgr from emp));
102.
List the employee names and his average salary department wise.
A)
select d.deptno, round(avg(nvl2(e1.comm, e1.sal+e1.comm, e1.sal))) avg, e2.ename from emp e1, emp e2, dept d where d.deptno =e1.deptno and d.deptno = e2.deptno group by d.deptno, e2.ename; (or)
B) select d.maxsal,e.ename,e.deptno as "current sal" from emp e,
(select avg(Sal) maxsal,deptno from emp group by deptno) d
where e.deptno=d.deptno;
103. Find out least 5 earners of the company.
A) s
elect * from emp e where 5> (select count(*) from emp where e.sal >sal); (or)
B)
select rownum rank,empno,ename,job,sal from (select * from emp order by sal asc) where rownum < 6 ; (or)
C)
select * from emp e where 5 >(select count(distinct sal) from emp where e.sal > sal);
104. Find out emps whose salaries greater than salaries of their managers.
A) s
elect * from emp w,emp m where w.mgr = m.empno and w.sal> m.sal; (OR)
B)
select * from emp e ,(select * from emp where empno in (select mgr from emp)) a
where e.sal >a.sal and e.mgr = a.empno
105. List the managers who are not working under the president.
A) select * from emp where empno in(select mgr from emp) and mgr not in
(select empno from emp where job = 'PRESIDENT')
106. List the records from emp whose deptno isnot in dept.
107. List the Name , Salary, Comm and Net Pay is more than any other employee.
A) Select e.ename,e.sal,e.comm,nvl2(comm,sal+comm,sal) NETPAY
from emp e
where nvl2(comm,sal+comm,sal) > any (select sal from emp where empno =e.empno) ;
108. List the Enames who are retiring after 31-Dec-89 the max Job period is 20Y.
A) s
elect ename from emp where add_months(hiredate,240) > '31-DEC-89';
B) select ename from emp
where add_months(hiredate,240) > to_date(’31-DEC-89’,’DD-MON-RR’);
109. List those Emps whose Salary is odd value.
A) s
elect * from emp where mod(sal,2) = 1;
110. List the emp’s whose Salary contain 3 digits.
A) s
elect * from emp where length (sal) = 3;
111. List the emps who joined in the month of DEC.
A) s
elect * from emp where to_char(hiredate,’MON’) =’DEC’; (OR)
B)
select * from emp where to_char(hiredate,’MON’) in (‘DEC’); (OR)
C)
select * from emp where to_char(hiredate,’MONTH’) like ‘DEC%’;
11
2. List the emps whose names contains ‘A’.
A) select * from emp where ename like ‘%A%’;
113. List the emps whose Deptno is available in his Salary.
A) select * from emp where instr(sal,deptno) > 0;
114. List the emps whose first 2 chars from Hiredate=last 2 characters of Salary.
A) select * from emp
where substr(hiredate,1,2) = substr(sal,length(sal)-1,length(sal));
115. List the emps Whose 10% of Salary is equal to year of joining.
A) select * from emp where to_char(hiredate,'YY') in (select .1*sal from emp);
116. List first 50% of chars of Ename in Lower Case and remaining are upper Case.
A)
select lower(substr(ename,1,round(length(ename)/2)))
||substr(ename,round(length(ename)/2)+1,length(ename)) from emp ; (OR)
B) select lower(substr(ename,1,ciel(length(ename)/2)))
|| substr(ename,ciel(length(ename)/2)+1,length(ename)) from emp ;
117. List the Dname whose No. of Emps is =to number of chars in the Dname.
A) s
elect * from dept d where length(dname) in (select count(*) from emp e where e.deptno = d.deptno ); (or)
B)
select d.dname,count(*) from emp e ,dept d where e.deptno = d.deptno group by d.dname having count(*) = length (d.dname);
118. List the emps those who joined in company before 15th of the month.
A) s
elect * from emp where to_char(hiredate,'DD') < '15';
119. List the Dname, no of chars of which is = no. of emp’s in any other Dept.
A) s
elect * from dept d where length(dname) in (select count(*) from emp where d.deptno <> deptno group by deptno ); (or)
B)
select * from dept where length(dname) = any (select count(*) from emp where d.deptno <> deptno group by deptno);
C)
select * from dept d , (select count(*) s,e.deptno "M"from emp e group by e.deptno) d1
where length(dname)=d1.s and d1.M <>d.deptno;
120. List the emps who are working as Managers.
A) s
elect * from where job = ‘MANAGER’; (or)
B)
select * from emp where empno in (select mgr from emp );
121. List THE Name of dept where highest no.of emps are working.
A) select dname from dept where deptno in
(select deptno from emp group by deptno
having count(*) in
(select max(count(*)) from emp group by deptno) );
122. Count the No.of emps who are working as ‘Managers’(using set option).
A)se
lect count(*)
from(select * from emp minus select * from emp where job != 'MANAGER')
123. List the emps who joined in the company on the same date.
A) s
elect * from emp e where hiredate in
(select hiredate from emp where e.empno <> empno);
124. List the details of the emps whose Grade is equal to one tenth of Sales Dept.
A) select * from emp e,salgrade s
where e.sal between s.losal and s.hisal and
s.grade = 0.1* (select deptno from dept where dname = 'SALES');
125. List the name of the dept where more than average no. of emps are working.
A) select d.dname from dept d, emp e where e.deptno = d.deptno
group by d.dname
having count(*) > (select avg(count(*)) from emp group by deptno);
126. List the Managers name who is having max no.of emps working under him.
A
)select m.ename,count(*) from emp w,emp m
where w.mgr = m.empno
group by m.ename
having count(*) = (select max(count(*)) from emp group by mgr);
(OR)
B) select * from emp where empno = (select mgr from emp group by mgr having count(*) = (select max(count(*)) from emp group by mgr)) ;
127. List the Ename and Sal is increased by 15% and expressed as no.of Dollars.
A) select ename,to_char(1.15*sal,'$99,999') as "SAL" from emp; (only for $ it works)
B) select ename,'$'||1.15*sal “SAL” from emp;
128. Produce the output of EMP table ‘EMP_AND_JOB’ for Ename and Job.
A) select ename|| job as "EMP_AND_JOB" from emp ;
129. Produce the following output from EMP.
EMP
LOYEE
SMITH (clerk)
ALLEN (Salesman)
A) select ename || ‘(‘|| lower(job)||’)’ as “EMPLOYEE” from emp;
130) List the emps with Hire date in format June 4, 1988.
A) s
elect empno,ename,sal, to_char(hiredate,'MONTH DD,YYYY') from emp;
13
1) Print a list of emp’s Listing ‘just salary’ if Salary is more than 1500, on target if Salary is 1500 and ‘Below 1500’ if Salary is less than 1500.
A) s
elect empno,ename,sal|| ‘JUST SALARY’ "SAL" from emp where sal > 1500 union
select empno,ename, sal|| ‘ON TARGET’ "SAL" from emp where sal = 1500
union
select empno,ename, sal|| ‘BELOW 1500’ "SAL" from emp where sal < 1500; (OR)
B)select empno,ename,sal,job,
case
when sal = 1500 then 'ON TARGET'
when sal < 1500 then 'BELOW 1500'
when sal > 1500 then 'JUST SALARY'
else 'nothing'
end "REVISED SALARY"
from emp;
132) Write a query which return the day of the week for any date entered in format ‘DD-MM-YY’.
A) s
elect to_char(to_date('& s','dd-mm-yy'),'day') from dual ;
133) Write a query to calculate the length of service of any employee with the company, use DEFINE to avoid repetitive typing of functions.
A) D
EFINE service = ((months_between(sysdate,hiredate))/12)
B)
Select empno,ename,&service from emp where ename = ‘& name’;
134) Give a string of format ‘NN/NN’, verify that the first and last two characters are numbers and that the middle character is’/’. Print the expression ‘YES’ if valid, ‘NO’ if not valid. Use the following values to test your solution. ‘12/34’,’01/1a’, ‘99/98’.
A)
135) Emps hired on or before 15th of any month are paid on the last Friday of that month those hired after 15th are paid on the first Friday of the following month. Print a list of emps their hire date and the first pay date. Sort on hire date.
A) s
elect ename,hiredate,next_day(last_day(hiredate),'FRIDAY')-7 from emp where to_char(hiredate,'DD') <=15
union
select ename,hiredate,next_day(last_day(hiredate),'FRIDAY') from emp where to_char(hiredate,'DD') > 15;
136) Count the no. of characters with out considering spaces for each name.
A) s
elect length(replace(ename,’ ‘,null)) from emp;
13
7) Find out the emps who are getting decimal value in their Sal without using like operator.
A) s
elect * from emp where instr(sal,’.’,1,1) > 0;
138) List those emps whose Salary contains first four digit of their Deptno.
A) s
elect * from emp where instr(to_char(sal,,9999),deptno,1,1)>0 and instr(to_char(sal,9999),deptno,1,2)> 0 ;
13
9) List those Managers who are getting less than his emps Salary.
A) s
elect distinct m.ename,m.sal from emp w,emp m where w.mgr = m.empno and w.sal>m.sal;
B)
select * from emp w where sal < any ( select sal from emp where w.empno=mgr);
C)
select * from emp w where empno in ( select mgr from emp where
w.sal<sal);
140) Print the details of all the emps who are sub-ordinates to Blake.
A) s
elect * from emp where mgr in (select empno from emp where ename = 'BLAKE');
14
1) List the emps who are working as Managers using co-related sub-query.
A) s
elect * from emp where empno in (select mgr from emp);
142) List the emps whose Mgr name is ‘Jones’ and also with his Manager name.
A) s
elect w.ename,m.ename,(select ename from emp where m.mgr = empno) "his MANAGER"
from emp w,emp m where w.mgr = m.empno and m.ename = 'JONES'; (or)
B) select e.ename,w.ename,m.ename from emp e,emp w,emp m where e.mgr = w.empno and w.ename = ‘JONES’ and w.mgr = m.empno;
143) Define a variable representing the expression used to calculate on emps total annual remuneration use the variable in a statement, which finds all emps who can earn 30000 a year or more.
A) Set define on
B)
Define annual = 12*nvl2(comm.,sal+comm.,sal) (here define variable is a session variable)
C)
Select * from emp where &annual > 30000;
144) Find out how may Managers are their in the company.
A) s
elect count(*) from emp where job = ‘MANAGER’; (or)
B)
select count(*) from emp where empno in (select mgr from emp); (or)
C)
select count(distinct m.empno) from emp w,emp m where w.mgr = m.empno ;
14
5) Find Average salary and Average total remuneration for each Job type. Remember Salesman earn commission.secommm
A) s
elect avg(sal),avg(sal+nvl(comm,0)) from emp;
146) Check whether all the emps numbers are indeed unique.
A) s
elect empno,count(*) from emp group by empno;
147) List the emps who are drawing less than 1000 Sort the output by Salary.
A)select * from emp where sal < 1000 order by sal;
148) List the employee Name, Job, Annual Salary, deptno, Dept name and grade who earn 36000 a year or who are not CLERKS.
A)se
lecte.ename,e.job,(12*e.sal)"ANNUALSALARY", e.deptno,d.dname,s.grade
from emp e,dept d ,salgrade s where e.deptno = d.deptno and e.sal between s.losal and s.hisal
and (((12*e.sal)>= 36000) or (e.job != 'CLERK'))
149) Find out the Job that was filled in the first half of 1983 and same job that was filled during the same period of 1984.
A) s
elect * from emp where (to_char(hiredate,'MM ') <= 06 and to_char(hiredate,'YYYY') = 1984) and job in (select job from emp where to_char(hiredate,'MM' ) <= 06 and to_char(hiredate,'YYYY') <= 1983) ;
150) Find out the emps who joined in the company before their Managers.
A) s
elect * from emp w,emp m where w.mgr = m.empno and
w.hiredate< m.hiredate;(or)
B) select * from emp e where hiredate < (select hiredate from emp where empno = e.mgr)
151) List all the emps by name and number along with their Manager’s name and number. Also List KING who has no ‘Manager’.
A) s
elect w.empno,w.ename,m.empno,m.ename from emp w,emp m where w.mgr= m.empno(+);
152) Find all the emps who earn the minimum Salary for each job wise in ascending order.
A) s
elect * from emp where sal in
(select min(sal) from emp group by job)
order by sal asc;
153) Find out all the emps who earn highest salary in each job type. Sort in descending salary order.
A) s
elect * from emp where sal in
(select max(sal) from emp group by job)
order by sal desc;
154) Find out the most recently hired emps in each Dept order by Hiredate.
A) s
elect * from emp e where hiredate in
(select max(hiredate) from emp where e.deptno = deptno )
order by hiredate;
155) List the employee name,Salary and Deptno for each employee who earns a salary greater than the average for their department order by Deptno.
A) s
elect * from emp e
where sal > (select avg(sal) from emp where e.deptno = deptno );
B) select e.ename,e.sal,e.deptno from emp e,(select avg(sal) A,deptno D from
emp group by deptno) D1 where D1.D = e.deptno and e.sal > D1.A;
156) List the Deptno where there are no emps.
A) s
elect deptno ,count(*) from emp
group by deptno
having count(*) = 0;
157) List the No.of emp’s and Avg salary within each department for each job.
A) s
elect count(*),avg(sal),deptno,job from emp
group by deptno,job;
158) Find the maximum average salary drawn for each job except for ‘President’.
A) s
elect max(avg(sal)) from emp where job != 'PRESIDENT' group by job;
159) Find the name and Job of the emps who earn Max salary and Commission.
A) s
elect * from emp where sal = (select max(sal) from emp) and comm. is not null;
160) List the Name, Job and Salary of the emps who are not belonging to the department 10 but who have the same job and Salary as the emps of dept 10.
A) s
elect ename,job,sal from emp where deptno != 10 and job in (select job from emp where deptno = 10)
and sal in (select sal from emp where deptno = 10);
161) List the Deptno, Name, Job, Salary and Sal+Comm of the SALESMAN who are earning maximum salary and commission in descending order.
A)se
lect deptno,name,job,sal,sal+nvl(comm.,0) from emp where job = ‘SALESMAN’ and sal in (select max(sal+nvl(comm.,0)) from emp where comm. is not null)
Order by (sal +nvl(comm.,0)) desc;
162) List the Deptno, Name, Job, Salary and Sal+Comm of the emps who earn the second highest earnings (sal + comm.).
A) s
elect deptno,ename,sal,job,sal+nvl(comm,0) from emp e where 2 = (select count(distinct sal+nvl(comm,0)) from emp where (e.sal+nvl(comm.,0))<(sal+nvl(comm.,0));
163) List the Deptno and their average salaries for dept with the average salary less than the averages for all department
A) select deptno,avg(sal) from emp group by deptno
ha
ving avg(sal) <(select avg(Sal) from emp);
164) List out the Names and Salaries of the emps along with their manager names and salaries for those emps who earn more salary than their Manager.
A) s
elect w.ename,w.sal,m.ename,m.sal from emp w,emp m
where w.mgr = m.empno and w.sal > m.sal;
165) List out the Name, Job, Salary of the emps in the department with the highest average salary.List out the Name, Job, Salary of emps in department with highest average salary.O
A) s
elect * from emp where deptno in
(select deptno from emp e
having avg(sal) =(select max(avg(sal)) from emp group by deptno)
group by deptno);
166) List the empno,sal,comm. Of emps.
A) s
elect empno,sal,comm. from emp;
167) List the details of the emps in the ascending order of the sal.
A) select * from emp order by sal asc;
168) List the dept in the ascending order of the job and the desc order of the emps print empno, ename.
A) s
elect * from emp e order by e.job asc,e.empno desc ;
169) Display the unique dept of the emps.
A)select * from dept where deptno in (select unique deptno from emp);
170) Display the unique dept with jobs.
A) s
elect unique deptno ,job from emp ;
171) Display the details of the blake.
A) s
elect * from emp where ename = ‘BLAKE’;
172) List all the clerks.
A) s
elect * from emp where job = ‘CLERK’;
173) list all the employees joined on 1st may 81.
A) s
elect * from emp where hiredate = ’01-MAY-81’;
174) List the empno,ename,sal,deptno of the dept 10 emps in the ascending order of salary.
A) s
elect e.empno,e.ename,e.sal,e.deptno from emp where e.deptno = 10
or
der by e.sal asc;
175) List the emps whose salaries are less than 3500.
A) s
elect * from emp where sal <3500;
176) List the empno,ename,sal of all the emp joined before 1 apr 81.
A) s
elect e.empno ,e.ename .e.sal from emp where hiredate <’01-APR-81’;
177) List the emp whose annual sal is <25000 in the asc order of the salaries.
A) s
elect * from emp where (12*sal) < 25000 order by sal asc;
178) List the empno,ename,annsal,dailysal of all the salesmen in the asc ann sal
A) s
elect e.empno,e.ename ,12*sal "ANN SAL" , (12*sal)/365 "DAILY SAL" from emp e
where e.job = 'SALESMAN'
order by "ANN SAL" asc ;
179) List the empno,ename,hiredate,current date & exp in the ascending order of the exp.
A) s
elect empno,ename,hiredate,(select sysdate from dual),((months_between(sysdate,hiredate))/12) EXP
from emp
order by EXP asc;
180) List the emps whose exp is more than 10 years.
A) select * from emp where ((months_between(sysdate,hiredate))/12) > 10;
181) List the empno,ename,sal,TA30%,DA 40%,HRA 50%,GROSS,LIC,PF,net,deduction,net allow and net sal in the ascending order of the net salary.
182) List the emps who are working as managers.
A) s
elect * from emp where job = ‘MANAGER’;
183) List the emps who are either clerks or managers.
A) s
elect * from emp where job in (‘CLERK’,’MANAGER’);
184) List the emps who have joined on the following dates 1 may 81,17 nov 81,30 dec 81
A) select * from emp where to_char(hiredate,’DD-MON-YY’) in
(’
01-MAY-81’,’17-NOV-81’,’30-DEC-81’);
185) List the emps who have joined in the year 1981.
A) select * from emp where to_char(hiredate,’YYYY’) = ‘1981’;
186) List the emps whose annual sal ranging from 23000 to 40000.
A) s
elect * from emp where (12* sal) between 23000 and 40000;
187) List the emps working under the mgrs 7369,7890,7654,7900.
A) select * from emp where mgr in ( 7369,7890,7654,7900);
188) List the emps who joined in the second half of 82.
A)select * from emp where hiredate between ’01-JUL-82’ and ’31-DEC-82’;
189) List all the 4char emps.
A) select * from emp where length (ename) = 4;
190) List the emp names starting with ‘M’ with 5 chars.
A) s
elect * from emp where ename like ‘M%’ and length (ename) = 5;
191) List the emps end with ‘H’ all together 5 chars.
A) s
elect * from emp where ename like ‘%H’ and length (ename) = 5;
192) List names start with ‘M’.
A) s
elect * from emp where ename like ‘M%’;
193) List the emps who joined in the year 81.
A) s
elect * from emp where to_char(hiredate,’YY’) = ‘81’;
194) List the emps whose sal is ending with 00.
A)
select * from where sal like ‘’;
195) List the emp who joined in the month of JAN.
A) s
elect * from emp where to_char(hiredate,’MON’) = ‘JAN’; (OR)
B)
select * from emp where to_char (hiredate,’MM’) = 1;
196) Who joined in the month having char ‘a’.
A) s
elect * from emp where to_char (hiredate,’MONTH’) like’%A%’; (OR)
B)
select * from emp where instr(to_char(hiredate,’MONTH’),’A’) >0;
197) Who joined in the month having second char ‘a’
A) s
elect * from emp where to_char(hiredate,’MON’) like ‘_A%’; (OR)
B)
select * from emp where instr(to_char(hiredate,’MON’),’A’) = 2;
198) List the emps whose salary is 4 digit number.
A) s
elect * from emp where length (sal) = 4;(OR)
B)
select * from emp where sal between 999 and 9999;
199) List the emp who joined in 80’s.
A) s
elect * from emp where to_char(hiredate,’YY’) between ‘80’ and ’89’; (OR)
B)
select * from emp where to_char(hiredate,’YY’) >= ‘80’ and to_char(hiredate,’YY’) < ‘90’;
200) List the emp who are clerks who have exp more than 8ys.
A) s
elect * from emp where job = ‘CLERK’ and (months_between(sysdate,hiredate) /12) > 8;
201) List the mgrs of dept 10 or 20.
A) s
elect * from emp where job = ‘MANAGER’ and (deptno = 10 or deptno =20);
202) List the emps joined in jan with salary ranging from 1500 to 4000.
A) s
elect * from emp where to_char(hiredate,’MON’) = ‘JAN’ and sal
be
tween 1500 and 4000;
203) List the unique jobs of dept 20 and 30 in desc order.
A) s
elect distinct job from emp where deptno in (20,30) order by job desc;
204) List the emps along with exp of those working under the mgr whose number is starting with 7 but should not have a 9 joined before 1983.
A) s
elect * from emp where (mgr like '7%' and mgr not like '%9%')
and to_char(hiredate,'YY') < '83';
205) List the emps who are working as either mgr or analyst with the salary ranging from 2000 to 5000 and with out comm.
A) s
elect * from emp where (job in (‘MANAGER’ ,’ANALYST’) ) and sal between 2000 and 5000 and comm is null;
206) List the empno,ename,sal,job of the emps with /ann sal <34000 but receiving some comm. Which should not be>sal and desg should be sales man working for dept 30.
A) s
elect empno,ename,sal,job from emp where
12*(sal+nvl(comm,0)) < 34000 and comm is not null and comm<sal and job = 'SALESMAN' and deptno = 30;
207) List the emps who are working for dept 10 or 20 with desgs as clerk or analyst with a sal is either 3 or 4 digits with an exp>8ys but does not belong to mons of mar,apr,sep and working for mgrs &no is not ending with 88 and 56.
A) s
elect * from emp where
deptno in (10,20) and
job in ('CLERK','ANALYST') and
(length(sal) in (3,4)) and
((months_between(sysdate,hiredate))/12)> 8 and
to_char(hiredate,'MON') not in ('MAR','SEP','APR') and
(mgr not like '%88' and mgr not like '%56');
208) List the empno,ename,sal,job,deptno&exp of all the emps belongs to dept 10 or 20 with an exp 6 to 10 y working under the same mgr with out comm. With a job not ending irrespective of the position with comm.>200 with exp>=7y and sal<2500 but not belongs to the month sep or nov workingunder the mgr whose no is not having digits either 9 or 0 in the asc dept& desc dept
A)
209) List the details of the emps working at Chicago.
A) s
elect * from emp where deptno in (select deptno from dept where dept.loc = ‘CHICAGO’);
210) List the empno,ename,deptno,loc of all the emps.
A) s
elect e.empno,e.ename,e.deptno,d.loc from emp e ,dept d
where e.deptno = d.deptno ;
211) List the empno,ename,loc,dname of all the depts.,10 and 20.
A) s
elect e.empno,e.ename,e.deptno,d.loc,d.dname from emp e ,dept d
where e.deptno = d.deptno and e.deptno in (10,20);
212) List the empno, ename, sal, loc of the emps working at Chicago dallas with an exp>6ys.
A) s
elect e.empno,e.ename,e.deptno,e.sal,d.loc from emp e ,dept d
where e.deptno = d.deptno and d.loc in ('CHICAGO','DALLAS')
and (months_between(sysdate,hiredate)/12)> 6 ;
213) List the emps along with loc of those who belongs to dallas ,newyork with sal ranging from 2000 to 5000 joined in 81.
A) s
elect e.empno,e.ename,e.deptno,e.sal,d.loc from emp e ,dept d
where e.deptno = d.deptno and d.loc in ('NEW YORK','DALLAS')
and to_char(e.hiredate,'YY') = '81' and e.sal between 2000 and 5000;
214) List the empno,ename,sal,grade of all emps.
A) s
elect e.empno,e.ename,e.sal,s.grade from emp e ,salgrade s
where e.sal between s.losal and s.hisal ;
215) List the grade 2 and 3 emp of Chicago.
A) s
elect * from emp where empno in
(select empno from emp e,salgrade s where e.sal between s.losal and
s.hisal and s.grade in (2,3));
216) List the emps with loc and grade of accounting dept or the locs dallas or Chicago with the grades 3 to 5 &exp >6y
A) s
elect e.deptno,e.empno,e.ename,e.sal,d.dname,d.loc,s.grade from emp e,salgrade s,dept d
wh
eree.deptno = d.deptno and e.sal between s.losal and s.hisal
and s.grade in (3,5)
and ((months_between(sysdate,hiredate))/12) > 6
and ( d.dname = 'ACCOUNTING' or D.loc in ('DALLAS','CHICAGO'))
217) List the grades 3 emps of research and operations depts.. joined after 1987 and whose names should not be either miller or allen.
A) s
elect e.ename from emp e ,dept d,salgrade s
where e.deptno = d.deptno and d.dname in ('OPERATIONS','RESEARCH') and e.sal between s.losal and s.hisal
and e.ename not in ('MILLER','ALLEN')
and to_char(hiredate,'YYYY') >1987;
218) List the emps whose job is same as smith.
A) s
elect * from emp where job = (select job from emp where ename = 'SMITH');
219) List the emps who are senior to mi
220) List the emps whose job is same as either allen or sal>allen.
A) s
elect * from emp
where job = (select job from emp where ename = 'ALLEN')
or sal > (select sal from emp where ename = 'ALLEN');
221) List the emps who are senior to their own manager.
A) s
elect * from emp w,emp m where w.mgr = m.empno and
w.hiredate < m.hiredate;
222) List the emps whose sal greater than blakes sal.
A) s
elect * from emp
where sal>(select sal from emp where ename = ‘BLAKE’);
223) List the dept 10 emps whose sal>allen sal.
A) s
elect * from emp where deptno = 10 and
sal > (select sal from emp where ename = 'ALLEN');
224) List the mgrs who are senior to king and who are junior to smith.
A)se
lect * from emp where empno in
(select mgr from emp
where hiredate<(select hiredate from emp where ename = 'KING' )
and hiredate > (select hiredate from emp where ename = 'SMITH')) and mgr is
not null;
225) List the empno,ename,loc,sal,dname,loc of the all the emps belonging to king dept.
A) s
elect e.empno,e.ename,d.loc,e.sal,d.dname from emp e,dept d
where e.deptno=d.deptno and e.deptno in
(select deptno from emp where ename = 'KING'and emp.empno <> e.empno);
226) List the emps whose salgrade are greater than the grade of miller.
A) s
elect * from emp e,salgrade s
where e.sal between s.losal and s.hisal and s.grade >
(select s.grade from emp e,salgrade s where e.sal between s.losal and s.hisal and e.ename = 'MILLER') ;
227) List the emps who are belonging dallas or Chicago with the grade same as adamsor exp more than smith.
A) s
elect * from emp e ,dept d,salgrade s
where e.deptno= d.deptno and d.loc in ('DALLAS','CHICAGO') and e.sal between s.losal and s.hisal and
(s.grade in (select s.grade from emp e,salgrade s where e.sal between s.losal and s.hisal and e.ename = 'ADAMS')
or months_between (sysdate,hiredate) > (select months_between(sysdate,hiredate) from emp where ename = 'SMITH')) ;
228) List the emps whose sal is same as ford or blake.
A) s
elect * from emp where sal in (select sal from emp e where e.ename in ('FORD','BLAKE')and emp.empno <> e.empno);
229) List the emps whose sal is same as any one of the following.
A) s
elect * from emp where sal in
(select sal from emp e where emp.empno <> e.empno);
230) Sal of any clerk of emp1 table.
A) s
elect * from emp where job = ‘CLERK’;
231) Any emp of emp2 joined before 82.
A) select * from emp where to_char(hiredate,'YYYY') < 1982;
232) The total remuneration (sal+comm.) of all sales person of Sales dept belonging to emp3 table.
A) s
elect * from emp e
where (sal+nvl(comm,0)) in
(select sal+nvl(comm,0) from emp e,dept d where e.deptno=d.deptno
and d.dname = 'SALES'and e.job = 'SALESMAN');
233) Any Grade 4 emps Sal of emp 4 table.
A) s
elect * from emp4 e,salgrade s where e.sal between s.losal and s.hisal and s.grade = 4;
234) Any emp Sal of emp5 table.
A) s
elect * from emp5;
235) List the highest paid emp.
A) s
elect * from emp where sal in (select max(sal) from emp);
236) List the details of most recently hired emp of dept 30.
A) s
elect * from emp where hiredate in
(select max(hiredate) from emp where deptno = 30);
237) List the highest paid emp of Chicago joined before the most recently hired emp of grade 2.
A) s
elect * from emp
where sal = ( select max(sal) from emp e,dept d where e.deptno =
d.deptno and d.loc = ‘CHICAGO’ and
hiredate <(select max(hiredate) from emp e ,salgrade s
where e.sal between s.losal and s.hisal and s.grade = 2))
238) List the highest paid emp working under king.
A)se
lect * from emp where sal in
(select max(sal) from emp where mgr in

(select empno from emp where ename = 'KING'));

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Wednesday, August 14, 2013

SQL Server Queries

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