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SQL数据库语法及题目汇总
问题描述:
SQL经典题目之一:求部门工资前三高的所有员工。
题目内容包含两个表格Employee和Department 。
#Employee
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+
#Department
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
答案整理:
解法一:使用窗口函数
解题思路:
1) 因为Employee中有DepartmentId,可以直接根据部门ID分组,给员工工资排序,选择前三员工信息;
2)Join连接Employee表和Department表,得出部门名称。
复习窗口函数知识点链接
SELECT
B.Name AS Department,
A.Name AS Employee,
A.Salary
FROM (SELECT DENSE_RANK() OVER (partition by DepartmentId order by Salary desc) AS ranking,DepartmentId,Name,Salary
FROM Employee) AS A
JOIN Department AS B ON A.DepartmentId=B.id
WHERE A.ranking<=3
#结果输出:
IT Joe 85000.00
IT Randy 85000.00
IT Will 70000.00
IT Max 69000.00
IT Janet 69000.00
Sales Henry 80000.00
Sales Same 60000.00
解法二:使用 JOIN 和Where
解题思路:公司里前 3 高的薪水意味着有不超过 3 个工资比这些值大。
SELECT
d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
FROM
Employee e1
JOIN
Department d ON e1.DepartmentId = d.Id
WHERE
#工资级别数量小于等于3,即最多只有3个工资级别,也就是前三高
3 > (SELECT
COUNT(DISTINCT e2.Salary)
FROM
Employee e2
WHERE
#e2的工资级别大于等于e1的工资级别
e2.Salary > e1.Salary
AND e1.DepartmentId = e2.DepartmentId
)
;
解法三:使用 JOIN 和Having
SELECT
d.NAME AS Department,
e1.NAME AS Employee,
e1.salary AS Salary
FROM
employee AS e1
LEFT JOIN employee AS e2 ON e1.DepartmentId = e2.DepartmentId
AND e1.Salary < e2.Salary
LEFT JOIN department d ON e1.DepartmentId = d.id
GROUP BY
e1.Id
HAVING
count( DISTINCT e2.Salary ) <= 2
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/department-top-three-salaries
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