• mysql分组排序


    来源LeetCode

    部门工资前三高的所有员工-被关联表

    Employee 表包含所有员工信息,每个员工有其对应的工号 Id,姓名 Name,工资 Salary 和部门编号 DepartmentId

    +----+-------+--------+--------------+
    | 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 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,找出每个部门获得前三高工资的所有员工。例如,根据上述给定的表,查询结果应返回:

    IT 部门中,Max 获得了最高的工资,Randy 和 Joe 都拿到了第二高的工资,Will 的工资排第三。销售部门(Sales)只有两名员工,Henry 的工资最高,Sam 的工资排第二。

    +------------+----------+--------+
    | Department | Employee | Salary |
    +------------+----------+--------+
    | IT         | Max      | 90000  |
    | IT         | Randy    | 85000  |
    | IT         | Joe      | 85000  |
    | IT         | Will     | 70000  |
    | Sales      | Henry    | 80000  |
    | Sales      | Sam      | 60000  |
    +------------+----------+--------+
    
    

    方法1-子查询

    
    SELECT
        d.Name AS 'Department', 
        e1.Name AS 'Employee', 
        e1.Salary
    FROM 
        Employee e1
    JOIN Department d ON e1.DepartmentId = d.Id
    WHERE
        3 > (SELECT
                COUNT(DISTINCT e2.Salary)
            FROM
                Employee e2
            WHERE
                e2.Salary > e1.Salary
                AND e1.DepartmentId = e2.DepartmentId
            )
    ;
    
    

    方法2-连接+group by + having

    通过左自连接求出每个部门排名前3的薪水;判断雇员们的薪水是否在上面的薪水中;

    // 写法1
    select 
        d.name as department, e1.name as employee, e1.salary as salary
    from 
        department d 
        join employee e1 on d.id = e1.departmentid
        join employee e2 on e1.departmentid = e2.departmentid and e1.salary<=e2.salary
    group by 
        d.name, e1.name
    having 
        count(distinct e2.salary)<=3
    order by 
        d.name, e1.salary desc
    
    
    

    // 写法2
    SELECT
      t3.`Name` Department,
      t1.`Name` Employee,
      t2.Salary
    FROM
      Employee t1
      INNER JOIN(
        SELECT
          e1.DepartmentId,
          e1.Salary
        FROM
          Employee e1
          LEFT JOIN Employee e2 ON e1.DepartmentId = e2.DepartmentId AND e1.Salary < e2.Salary
        GROUP BY
          e1.DepartmentId,
          e1.Salary
        HAVING
          COUNT(DISTINCT e2.Salary) <= 2
      ) t2 ON t1.DepartmentId = t2.DepartmentId AND t1.Salary = t2.Salary
      INNER JOIN Department t3 ON t1.DepartmentId = t3.Id
    
    
    

    方法3-变量

    1、对各个部门员工工资进行排序。核心思想是根据DepartmentId升序,Salary降序对Employee表进行排序,添加rank字段记录排名,组成临时表。rank字段规则:若上一条记录的DepartmentId与当前记录不同,则rank字段为1(不同部门);

    若上一条记录的DepartmentId、Salary均与当前记录相同,则rank字段保持不变(排名相同);

    若上一条记录的DepartmentId与当前记录相同,rank与当前记录不同,则SalaryRank字段+1。

    2、取出排名中名次小于等于三的员工

    3、与部门名称表进行连接

    SELECT 	
       d.NAME department, t.NAME employee, salary 
    FROM
       ( SELECT 
          *, @r := IF(@pD = departmentid, IF(@pS = salary, @r, @r + 1 ), 1 ) AS 'rank',
          @pD := departmentid,
          @pS := salary 
         FROM 
          employee, ( SELECT @pS := NULL, @pD := NULL, @r := 0 ) as init 
         ORDER BY
          departmentid, salary DESC ) t
       JOIN department d ON t.departmentid = d.id 
    WHERE
       t.rank <=3
    
    

    SELECT 
    max(ID) as risk_warning_id
    , related_guarantee_id
    FROM cfbiz_risk_warning 
    -- ifnull 是为了历史数据
    where 
    delete_flag = 0 and 
    ifnull(risk_warning_confirm_date, CREATED_TIME) <= curdate() 
    GROUP BY related_guarantee_id
    
    

    方法4-limit

    SELECT 
      d.Name as Department,
      e.Name as Employee,
      e.Salary as Salary
    FROM 
      Employee as e LEFT JOIN (
        SELECT 
          *, IFNULL((
            SELECT DISTINCT e1.Salary 
            FROM Employee as e1 
            WHERE e1.DepartmentId = Department.Id
            ORDER BY e1.Salary DESC
            LIMIT 2,1 -- 偏移,条数
          ),0) as TopSalary 
        FROM Department 
      ) as d ON e.DepartmentId = d.Id
    WHERE 
      e.Salary >= d.TopSalary
    
    

    方法5-窗口函数

    首先通过窗口函数dense_rank实现每个部门内部的工资排序,然后将员工的id和自己所在部门工资的名词进行绑定生成表c,然后三个表连接是,将排名是1,2,3的行选出来即可。

    select 
    d.Name as Department, 
    e2.Name as Employee, 
    e2.Salary 
    from 
    Department d inner join 
    (
        select e.*, 
        dense_rank() over(partition by DepartmentID 
        Order by Salary DESC) as 'rank'
        from Employee e 
    ) e2 on d.Id= e2.DepartmentID
    where e2.rank<=3
    order by Department AND Salary
    
    

    如果还有其他方法,不吝赐教

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  • 原文地址:https://www.cnblogs.com/mspeer/p/14056157.html
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