2020.7.8 LeetCode 从零单刷个人笔记整理(持续更新)
github:https://github.com/ChopinXBP/LeetCode-Babel
传送门:部门工资最高的员工
Employee 表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询,找出每个部门工资最高的员工。例如,根据上述给定的表格,Max 在 IT 部门有最高工资,Henry 在 Sales 部门有最高工资。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
#注意:1.最高工资不止一人;2.可能有人的部门号没有对应部门名
#左连接后要判断NULL值
SELECT d.Name AS 'Department', t.Name AS 'Employee', t.Salary AS 'Salary'
FROM (
SELECT e.Name, e.Salary, e.DepartmentId
FROM (SELECT DepartmentId, MAX(Salary) AS 'maxSalary'
FROM Employee
GROUP BY DepartmentId) m, Employee e
WHERE e.Salary = m.maxSalary AND e.DepartmentId = m.DepartmentId
) t
LEFT JOIN Department d
ON t.DepartmentId = d.Id
WHERE d.Name IS NOT NULL;
#内连接会自动去除NULL值
SELECT d.Name AS 'Department', t.Name AS 'Employee', t.Salary AS 'Salary'
FROM (
SELECT e.Name, e.Salary, e.DepartmentId
FROM (SELECT DepartmentId, MAX(Salary) AS 'maxSalary'
FROM Employee
GROUP BY DepartmentId) m, Employee e
WHERE e.Salary = m.maxSalary AND e.DepartmentId = m.DepartmentId
) t
INNER JOIN Department d
ON t.DepartmentId = d.Id;
#IN左右可以连接多个值对
SELECT d.Name AS 'Department', e.Name AS 'Employee', e.Salary AS 'Salary'
FROM Employee e
INNER JOIN Department d
ON e.DepartmentId = d.Id
WHERE (e.DepartmentId, e.Salary) IN (
SELECT DepartmentId, MAX(Salary)
FROM Employee
GROUP BY DepartmentId
);
#Coding一小时,Copying一秒钟。留个言点个赞呗,谢谢你#