leetcode181. 超过经理收入的员工(SQL)
Employee 表包含所有员工,他们的经理也属于员工。每个员工都有一个 Id,此外还有一列对应员工的经理的 Id。
+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+
给定 Employee 表,编写一个 SQL 查询,该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中,Joe 是唯一一个收入超过他的经理的员工。
+----------+
| Employee |
+----------+
| Joe |
+----------+
思路:
自连接
-
# Write your MySQL query statement below
-
select A.Name as 'Employee'
-
from Employee as A,Employee as B
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where A.ManagerId=B.Id and A.Salary>B.Salary;
子查询包含主查询内容的效率比较慢,所以不推荐使用。
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select a.name as Employee
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from Employee as a
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where a.salary > (select b.salary
-
from Employee as b
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where b.id = a.managerid);
文章来源: fantianzuo.blog.csdn.net,作者:兔老大RabbitMQ,版权归原作者所有,如需转载,请联系作者。
原文链接:fantianzuo.blog.csdn.net/article/details/104315225
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