时间：2021-12-31 09:07:25 | 栏目：Mysql | 点击：次
Write a SQL query to get the nth highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.
这道题是之前那道Second Highest Salary的拓展，根据之前那道题的做法，我们可以很容易的将其推展为N，根据对Second Highest Salary中解法一的分析，我们只需要将OFFSET后面的1改为N-1就行了，但是这样MySQL会报错，估计不支持运算，那么我们可以在前面加一个SET N = N - 1，将N先变成N-1再做也是一样的：
解法一：
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N = N - 1;
RETURN (
SELECT DISTINCT Salary FROM Employee GROUP BY Salary
ORDER BY Salary DESC LIMIT 1 OFFSET N
);
END
根据对Second Highest Salary中解法四的分析，我们只需要将其1改为N-1即可，这里却支持N-1的计算，参见代码如下：
解法二：
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
RETURN (
SELECT MAX(Salary) FROM Employee E1
WHERE N - 1 =
(SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary > E1.Salary)
);
END
当然我们也可以通过将最后的>改为>=，这样我们就可以将N-1换成N了：
解法三：
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
RETURN (
SELECT MAX(Salary) FROM Employee E1
WHERE N =
(SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary >= E1.Salary)
);
END
类似题目：
参考资料：
https://leetcode.com/discuss/88875/simple-answer-with-limit-and-offset
https://leetcode.com/discuss/63183/fastest-solution-without-using-order-declaring-variables