Finding Nth highest salary in a table is the most common question asked in interviews. Here is a way to do this task using dense_rank() function.
Consider the following table:
select * from( select ename, sal, dense_rank() over(order by sal desc)r from Employee) where r=&n; To find to the 2nd highest sal set n = 2 To find 3rd highest sal set n = 3 and so on.
1. DENSE_RANK computes the rank of a row in an ordered group of rows and returns the rank as a NUMBER. The ranks are consecutive integers beginning with 1.
2. This function accepts arguments as any numeric data type and returns NUMBER.
3. As an analytic function, DENSE_RANK computes the rank of each row returned from a query with respect to the other rows, based on the values of the value_exprs in the order_by_clause.
4. In the above query the rank is returned based on sal of the employee table. In case of tie, it assigns equal rank to all the rows.
Alternate Solution :
CREATE TABLE `Employee` (
`ENAME` varchar(225) COLLATE utf8_unicode_ci NOT NULL,
`SAL` bigint(20) unsigned NOT NULL,
PRIMARY KEY (`ENAME`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
6th highest mysql> select * from ((select * from Employee ORDER BY `sal` DESC limit 6 ) AS T) ORDER BY T.`sal` ASC limit 1; Alternate use of Limit: select * from Employee ORDER BY `sal` DESC limit 5,1; // will return 6th highest +-------+-----+ | ENAME | SAL | +-------+-----+ | B | 300 | +-------+-----+ 1 row in set (0.00 sec)
mysql> select * from Employee; +-------+-----+ | ENAME | SAL | +-------+-----+ | A | 100 | | B | 300 | | C | 200 | | D | 500 | | F | 400 | | G | 600 | | H | 700 | | I | 800 | +-------+-----+ 8 rows in set (0.00 sec)
Thanks to Vijay for suggesting this alternate solution.
Suppose the task is to find the employee with the Nth highest salary from the above table. We can do this as follows:
- Find the employees with top N distinct salaries.
- Find the lowest salary among the salaries fetched by the above query, this will give us the Nth highest salary.
- Find the details of the employee whose salary is the lowest salary fetched by the above query.
SELECT * FROM Employee WHERE sal = ( SELECT MIN(sal) FROM Employee WHERE sal IN ( SELECT DISTINCT TOP N sal FROM Employee ORDER BY sal DESC ) )
The above query will fetch the details of the employee with the Nth highest salary. Let us see how:
- Consider N = 4.
- Starting with the most inner query, the query: “SELECT DISTINCT TOP 4 sal FROM Employee ORDER BY sal DESC” will produce the below result:
51000 39800 35000 31500
The next outer query is: “SELECT MIN(sal) FROM Employee WHERE sal IN ( Result_Set_of_Previous_Query )“. This will return the below result:
You can see that the above returned result is the required 4th highest salary.
Next is the most outer query, which is: “SELECT * FROM Employee WHERE sal = Result_of_Previous_Query“. This query will return the details of employees with 4th highest salary.
________________________ ename sal ________________________ F | 31500 | ________________________
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Improved By : real1107055