About Level Psuedo column

Question

Hi all,
can any one explain the query how it is working internally.
I could not get the logic of it and i tried and surfed many sites, all the sites are giving the query but no one explained how it is working.
how will you get the nth highest salary ?
select * from emp where sal in (select max(sal) from emp where level = &level connect by prior sal > sal group by level);
Please help me.
Regards,
Sri Ram.

Answer 1

Hi,
Sri Ram wrote:
i want to know how the query is getting executed.If a row is in the result set because it meets the START WITH condition, then it has LEVEL = 1.
If a row is in the result set becuse it (together with some row that has LEVEL = N) meets the CONNECT BY condition, then the new row has LEVEL = N + 1.
If there is no START WITH clause, then every row in the table is counted as meeting the START WITH conditiion, and so every row in the table is in the result set once with LEVEL = 1.
Let's simplify the problem a little by including only deptno=20. That is, let's only work with the rows produced by:
CREATE OR REPLACE VIEW  emp_20  AS
SELECT       ename, sal
FROM       scott.emp
WHERE       deptno     = 20;so the data is:
ENAME             SAL
SMITH             800
ADAMS            1100
JONES            2975
FORD             3000
SCOTT            3000To understand any query, look at its sub-queries and make sure you understand them first. Run just the sub-query, simplified, if necessary, and showing any extra expressions that will help you understand it. In this case, the quesry contains the following sub-query:
     select        max(sal)
     from        emp_20
     where        level = &level_wanted
     connect by      prior sal > sal
     group by  level
)That's a lot to understand a once. Let's simplify it, by removing the GROUP BY. Of course, that means we won't be able to use aggregate functions like MAX, so let's display their arguments, instead. Also, let's start by not using a WHERE clause. Instead, let's display the expression used in the WHERE clause. When trying to understand a CONNECT BY query, it can be very helpful to include SYS_CONNECT_BY_PATH in the output.
So let's run just this query:
WITH     in_subquery     AS
     select        sal          -- originally was max (sal)
     ,       ename
     ,       LEVEL
     ,       SYS_CONNECT_BY_PATH (ename, ',')     AS path
     from        emp_20     -- originally was xcott.emp
--     where        level = &level_wanted
     connect by      prior sal > sal
--     group by  level
SELECT     *
FROM     in_subquery
;Output:
  SAL ENAME  LEVEL PATH
  800 SMITH      1 ,SMITH
2975 JONES      1 ,JONES
  800 SMITH      2 ,JONES,SMITH
1100 ADAMS      2 ,JONES,ADAMS
  800 SMITH      3 ,JONES,ADAMS,SMITH
3000 SCOTT      1 ,SCOTT
  800 SMITH      2 ,SCOTT,SMITH
2975 JONES      2 ,SCOTT,JONES
  800 SMITH      3 ,SCOTT,JONES,SMITH
1100 ADAMS      3 ,SCOTT,JONES,ADAMS
  800 SMITH      4 ,SCOTT,JONES,ADAMS,SMITH
1100 ADAMS      2 ,SCOTT,ADAMS
  800 SMITH      3 ,SCOTT,ADAMS,SMITH
1100 ADAMS      1 ,ADAMS
  800 SMITH      2 ,ADAMS,SMITH
3000 FORD       1 ,FORD
  800 SMITH      2 ,FORD,SMITH
2975 JONES      2 ,FORD,JONES
  800 SMITH      3 ,FORD,JONES,SMITH
1100 ADAMS      3 ,FORD,JONES,ADAMS
  800 SMITH      4 ,FORD,JONES,ADAMS,SMITH
1100 ADAMS      2 ,FORD,ADAMS
  800 SMITH      3 ,FORD,ADAMS,SMITHThere is no START WITH clause, so every employee appears once on LEVEL=1.
The CONNECT BY condition is "prior sal > sal", meaning that row p will be considered a parent of row c when p.sal > c.sal.
Let's look at the employee named JONES. the result set has the following rows with ename='JONES':
  SAL ENAME  LEVEL PATH
2975 JONES      1 ,JONES
2975 JONES      2 ,SCOTT,JONES
2975 JONES      2 ,FORD,JONESWe see that JONES is in the result set once with LEVEL=1(becuase there is no START WITH clause). JONES also appears as the child of SCOTT (because SCOTT has a higher sal than JONES) and again as a child of FORD (for the same reason). FORD was in the result set with LEVEL=1, and a row for JONES joined the result set at then next highest LEVEL, becuase that row met the CONNECT BY condition: FORD's sal (3000) was higher than JONES's sal (2975). Looking at this output, we can see that there is a chain of employees, each with a sal greater than the next employee. JONES appears in such a chain with LEVEL=2, so JONES has the 2nd highest sal in that chain, or JONES's sal (2975) is the 2nd highest sal in that chain.
Look at another employee, say ADAMS. Do you understand why all the rows with ename='ADAMS' are there, and what all the columns in the output mean on those rows?
When you understand this sub-query, let's modify it, making it a little more like the original. Let's restore the WHERE clause, so the query is now:
WITH     in_subquery     AS
     select        sal          -- originally was max (sal)
     ,       ename
     ,       LEVEL
     ,       SYS_CONNECT_BY_PATH (ename, ',')     AS path
     from        emp_20     -- originally was xcott.emp
     where        level = &level_wanted
     connect by      prior sal > sal
--     group by  level
SELECT     *
FROM     in_subquery
;and the output for any given value of &level_wanted will be something like:
  SAL ENAME  LEVEL PATH
  800 SMITH      2 ,JONES,SMITH
1100 ADAMS      2 ,JONES,ADAMS
  800 SMITH      2 ,SCOTT,SMITH
2975 JONES      2 ,SCOTT,JONES
1100 ADAMS      2 ,SCOTT,ADAMS
  800 SMITH      2 ,ADAMS,SMITH
  800 SMITH      2 ,FORD,SMITH
2975 JONES      2 ,FORD,JONES
1100 ADAMS      2 ,FORD,ADAMSAgain, study the output, and make sure you understand it.
Do you see jow each sal shown is the 2nd highest in some parent-child chain?
Do you see how heach sal shown might possibly be the 2nd highest sal?
When you do uderstand that query, make it still more like the original. Restore the GROUP BY clause and the MAX function. Actually, all you need is the MAX function. Since the WHERE clause guarantees that all rows in the ungrouped result set have the save LEVEL, then saying "GROUP BY LEVEL" isn't doing anything, and it doesn't make any sense. As Whitehat pointed out, the whole approach doesn't make any sense. This is a complicated and inefficient way to get the results you want. Whitehat showed a much better way, however, that solution was for the N-th highest paid employee. When there are ties (mutliple employees with the same sal), this will not be the same as the employee(s) who have the N-th highest sal, which is what the query you posted shows. Here's a better way to do what your query is doing:
WITH     got_r_num     AS
     SELECT     emp.*
     ,     DENSE_RANK () OVER (ORDER BY  sal   DESC)     AS r_num
     FROM     scott.emp
SELECT     ename, sal     -- list all columns wanted
FROM     got_r_num
WHERE     r_num     = &level_wanted
;