SQL Queries

Excellence through learning

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;