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 3Rd and Oct 4th 1994.
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 3rd.
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 3rd 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 6th.
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.city = ‘London’)
and city = ‘London’;
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);
90. Insert into table emp1 empno, sal and deptno from emp table.
If table emp1 is created then
insert into emp1 ( select empno,sal,deptno
from emp);
IF table is not created then
Create table emp1 as ( select empno,sal,deptno
from emp);
91. Update Salary of all employees by 10%.
Update emp
set sal = sal + 0.10 * sal;
92. Delete all rows from emp for deptno = 10.
Delete from emp
where deptno = 10;
93. Select list of all jobs which have an annual average salary greater than that managers.
Select job,avg(sal)
from emp
group by job
having avg(sal) > ( select avg(sal)
from emp
where job = ‘MANAGER’);
94. Select list of all employees who have atleast one other employee reporting to them.
Select a.job, a.ename, a.empno, a.deptno
from emp a
where exists ( select *
from emp b
where a.empno = b.mgr);
95. Select all employees with correponding level numbers.
Column orgn_chart format a21
Select lpad(‘ ‘,3*level)|| ename orgn_charts,level,empno,job,mgr
from emp
connect by prior empno = mgr
start with name = ‘KING’;
96. Select average salary for employee at each level.
Select level,avg(sal)
from emp
connect by prior empno = mgr
start with name = ‘KING’
group by level
order by level;
97. Display organization chart for only those employee who work under ‘JONES’.
Column orgn_chart format a21
Select lpad(‘ ‘,3*level)|| ename orgn_charts,level,empno,job,mgr
from emp
connect by prior empno = mgr
start with name = ‘JONES’;
98. Display organization chart for only those employee who work under ‘JONES’ and ‘BLAKE’.
Column orgn_chart format a21
Select lpad(‘ ‘,3*level)|| ename orgn_charts,level,empno,job,mgr
from emp
connect by prior empno = mgr
start with name in (‘JONES’,’BLAKE’);
99. List information about all the people in the organization above ‘ADAMS’.
Column orgn_chart format a21
Select lpad(‘ ‘,3*level)|| ename orgn_charts,empno,job,mgr
from emp
connect by empno = prior mgr
start with name = ‘ADAMS’;
100. List all the people who work under ‘BLAKE’ except ‘JAMES’.
Column orgn_chart format a21
Select lpad(‘ ‘,3*level)|| ename orgn_chart,level,empno,job,mgr
from emp
where ename != ‘JAMES’
connect by prior empno = mgr
start with ename = ‘BLAKE’;
101. List all the people who work under ‘KING’ except all employees
working under ‘BLAKE’.
Select lpad(‘ ‘,3*level)|| ename orgn_chart,level,empno,job,mgr
from emp
connect by prior empno = mgr
and ename != ‘BLAKE’
start with ename = ‘KING’
102. List all the people who work under ‘KING’ except ‘ADAMS’ and ‘BLAKE’ and all employees working under ‘BLAKE’.
Select lpad(‘ ‘,3*level)|| ename orgn_chart,level,empno,job,mgr
from emp
where ename != ‘ADAMS’
connect by prior empno = mgr
and ename != ‘BLAKE’
start with ename = ‘KING’
103. Select max salarys of deptno 10,20 and 30 in single row.
Select min(decode(deptno,10,max(sal))) “Dept No 10”,
min(decode(deptno,20,max(sal))) “Dept No 20”,
min(decode(deptno,30,max(sal))) “Dept No 30”
from emp
group by deptno;
104. If supply table has three fields vendor, job, part. Find list of vendor who
are supplying all part for given job.
Select a.vendor,a.job,count(*)
from supply a
group by a.vendor,a.job
having count(*) = ( select count(*)
from supply b
where a.job = b.job);
105. List all pairs of orders having same item and qty.
Select a.ordid, b.ordid
from item a, item b
where not exists ( select itemid,qty
from item c
where c.ordid = a.ordid
minus
select itemid,qty
from item d
where d.ordid = b.ordid ) and
not exists ( select itemid,qty
from item c
where c.ordid = b.ordid
minus
select itemid,qty
from item d
where d.ordid = a.ordid ) and
a.ordid < b.ordid
group by a.ordid,b.ordid
Table name: lot_mas t
Structure:
lot_no lot_desc storage
a Pentium fabrication
a 486 fabrication
b Pentium fabrication
d Pentium fabrication
d Pentium fabrication
. . .
. . .
. . .
I want to display the out put like
lot_no times
a 2
d 3
. .
. .
select lot_no,count(lot_no)
from lot_mast
group by lot_no
having count(lot_no) > 1;
There are twelve records for each employee in a year. The table structure
is as follow.
Empno month_no salary
1000 01 2000
1000 02 3500
1000 03 2500
…. .. ….
…. .. ….
1000 12 3450
2000 01 1900
2000 02 5000
…. .. ….
…. .. ….
2000 12 5450
3000 01 1900
3000 02 5000
…. .. ….
…. .. ….
The output should be as follows
empno Jan Feb Mar Apr May …… Dec
1000 2000 3500 2500 …. …. …… 3450
2000 1900 5000 …. …. …. …… 5450
3000 1900 5000 …. …. …. …… …….
…. …. …. …. …. …. …. ….
…. …. …. …. …. …. …. ….
…. …. …. …. …. …. …. ….
select a.empno,a.salary,b.salary,c.salary,d.salary,e.salary,
f.salary,g.salary,h.salary,i.salary,j.salary,k.salary,l.salary
from emp a,emp b,emp c,emp d,emp e,emp
f,emp g,emp h,emp i,emp j,emp,k.emp,emp l
where a.month = 1 and a.empno = b.empno and b.month = 2
and b.empno = c.empno and c.month = 3
and c.empno = d.empno and d.month = 4
and d.empno = e.empno and e.month = 5
and d.empno = f.empno and f.month = 6
and d.empno = g.empno and g.month = 7
and d.empno = h.empno and h.month = 8
and d.empno = i.empno and i.month = 9
and d.empno = j.empno and j.month = 10
and d.empno = k.empno and k.month = 11
and d.empno = l.empno and l.month = 12;
The table structure is as follows
start_site_name end_site_name
F1 F2
F2 F3
F3 F4
F5 F6
F6 F7
Fa Fb
Fb Fc
.. ..
.. ..
The out will be
level start_site_name end_site_name
1 F1 F2
2 F2 F3
3 F3 F4
1 F5 F6
2 F6 F7
1 Fa Fb
2 Fb Fc
.. .. ..
.. .. ..
.. .. ..
select level,start_site_name,end_site_name
from table_name
connect by prior end_site_name = start_site_name
start with start_site_name = F1;