以下查询在 MySql 中不起作用。
SELECT * /*This is the outer query part */
FROM Employee Emp1
WHERE (N-1) = ( /* Subquery starts here */
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
帮帮我。
最佳答案
试试这个:
SET @n := 0;
SELECT * FROM
(SELECT salary, @n := @n + 1 AS t1 FROM Employee ORDER BY salary DESC)
AS t
WHERE t1 <= 5
关于mysql - 如何在不使用限制的情况下在MySql中找到前5名薪水,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33252604/