“How can I select the Nth highest salary of the EMP table?”
This is a question that every Oracle newbie stumbles over. Ask it on a forum and you’re pointed to the archives. That gets you nowhere as when you search the archives, all you find is a host of other messages also asking you to search the archives.
Here comes the answer to the problem of finding the Nth highest salary. You can extend this logic to find the Nth highest row of any table.
First things first: The question is ambiguous!
Let’s say this is your data:
| Name | Salary |
| KING | 5000 |
| FORD | 3000 |
| SCOTT | 3000 |
| JONES | 2975 |
| BLAKE | 2850 |
| CLARK | 2850 |
| ALLEN | 1600 |
Who is second – FORD or SCOTT or both?
What will you say about JONES’s salary – is it the 3rd highest salary, or the 4th highest?
If you are looking for the set of people earning the Nth highest salary, with no gaps in case of ties, then JONES should be ranked 3rd, after KING [5000, 1st], followed by FORD and SCOTT [both 3000, 2nd].
If you are looking for exact ranks with gaps if there are ties, then JONES is the 4th highest paid employee, as there are 3 people earning more than him – KING, FORD and SCOTT. In this system of ranking, FORD and SCOTT are 2nd jointly and no employee is 3rd.
This is how your ranks will look, in the 2 cases:
|
Scenario 1: No gaps in case of ties Scenario 2: Gaps in case of ties
Once you have your question sorted out –
(a) Set of people earning the Nth highest salary, with continuous ranks if there are ties, OR
(b) Set of people earning the Nth highest salary, with skipped rank numbers if there are ties
Then you can proceed to writing the queries.
Scenario 1: DENSE_RANK () for Nth highest row, no gaps in case of ties
The analytic function dense_rank() will rank the rows with no gaps in ranking sequence if there are ties.
The ranks are calculated as:
SQL> select ename 2 ,sal 3 ,dense_rank() over (order by sal desc) ranking 4 from emp; ENAME SAL RANKING ---------- ---------- ---------- KING 5000 1 FORD 3000 2 SCOTT 3000 2 JONES 2975 3 CLARK 2850 4 BLAKE 2850 4 ALLEN 1600 5
Wrap a filter around and pick out the Nth highest salary, say the 4th highest salary.
SQL> select * 2 from 3 ( 4 select ename 5 ,sal 6 ,dense_rank() over (order by sal desc) ranking 7 from emp 8 ) 9 where ranking = 4 -- Replace 4 with any value of N 10 / ENAME SAL RANKING ---------- ---------- ---------- BLAKE 2850 4 CLARK 2850 4
The 4th position has a tie between BLAKE and CLARK.
Scenario 2: RANK () for Nth highest row, gaps in case of ties
The analytic function rank() will rank the rows with gaps in ranking sequence if there are ties.
The ranks are calculated as:
SQL> select ename 2 ,sal 3 ,rank() over (order by sal desc) ranking 4 from emp; ENAME SAL RANKING ---------- ---------- ---------- KING 5000 1 FORD 3000 2 SCOTT 3000 2 JONES 2975 4 CLARK 2850 5 BLAKE 2850 5 ALLEN 1600 7 TURNER 1500 8
Wrap a filter around and pick out the Nth highest salary, say the 4th highest salary.
SQL> select * 2 from 3 ( 4 select ename 5 ,sal 6 ,rank() over (order by sal desc) ranking 7 from emp 8 ) 9 where ranking = 4 -- Replace 4 with any value of N 10 / ENAME SAL RANKING ---------- ---------- ---------- JONES 2975 4
A different answer from the previous query, as there is no rank 3 because of the tied 2nd place.
Closing Notes
The requirement to “find Nth highest row” is incomplete, until the following questions are also answered:
- Can the result match more than one value? If not, on what basis should the one record be chosen if there is a tie?
- How should the subsequent records be ranked in case of ties – contiguously or with gaps?
Depending on the answer for (2), DENSE_RANK (for contiguous) or RANK (for gaps) can be used. Depending on the answer for (1), extra filter criteria can be applied to the SQL.
There are other approaches for calculating the Nth highest row, too. The next is a non-analytic approach, which works the same way as the RANK query (gaps for ties).
SQL> select ename 2 , sal 3 from emp a 4 where 3 = ( select count(*) -- Replace 3 with any value of (N - 1) 5 from emp b 6 where b.sal > a.sal) 7 / ENAME SAL ---------- ---------- JONES 2975
However, tests have shown the analytics approach to be more efficient than the non-analytics one for Nth highest or Top-N type of queries.
