找到第二高的薪水[重复]

Question

这是一个众所周知的问题。考虑以下

EmployeeID  EmployeeName    Department      Salary   
----------- --------------- --------------- ---------
1           T Cook          Finance         40000.00
2           D Michael       Finance         25000.00
3           A Smith         Finance         25000.00
4           D Adams         Finance         15000.00
5           M Williams      IT              80000.00
6           D Jones         IT              40000.00
7           J Miller        IT              50000.00
8           L Lewis         IT              50000.00
9           A Anderson      Back-Office     25000.00
10          S Martin        Back-Office     15000.00
11          J Garcia        Back-Office     15000.00
12          T Clerk         Back-Office     10000.00

我们需要找出第二高的薪水

With Cte As
(
  Select 
    level
    ,Department
    ,Max(Salary) 
 From plc2_employees
 Where level = 2
 Connect By Prior (Salary) > Salary)
Group By level,Department
)

Select 
    Employeeid
    ,EmployeeName
    ,Department
    ,Salary
From plc2_employees e1
Inner Join Cte e2 On e1.Department = e2.Department
Order By 
    e1.Department
    , e1.Salary desc
    ,e1.EmployeeID

不知何故无法正常工作...我没有得到正确的结果。谁能帮帮我。

Answer 1

类似

select * from
(
select EmployeeID, EmployeeName, Department, Salary, 
rank () over (partition by Department order by Salary desc) r
from PLC2_Employees
)
where r = 2

编辑 - 对其进行了测试，它给出了您期望的答案。

Answer 2

如果您要自学如何处理 CONNECT BY，则应首先找到适合该构造的问题。 CONNECT BY 用于处理分层形式的数据，而您的示例不是。薪金之间没有等级关系。试图在错误的问题上强制拟合一个结构是令人沮丧的，而且并不能真正教给你任何东西。

查看演示 HR 模式中的经典员工-经理关系，您可以使用 Oracle 安装。所有员工都向经理报告，包括经理（最高层除外）。然后，您可以使用此架构创建查询以显示例如公司的组织结构图。

Answer 3

START WITH … CONNECT BY 旨在通过探索所有可能的下降路径来探索形成图形的数据。您在START WITH 子句中指定根节点在CONNECT BY 子句中指定节点连接（不在WHERE 子句中）。

WHERE 子句过滤器将在 分层条件之后处理，GROUP BY 和 HAVING 相同（当然因为 GROUP BY 是在 WHERE 之后计算的）。

因此，例如，您必须在这里CONNECT BY PRIOR department = department。还必须避免在有中间工资的情况下，两个工资之间做节点连接。

因此，最终查询将类似于：

SELECT level
     , Department
     , Salary
  FROM plc2_employees pe1
 START WITH pe1.salary = (select max(salary) from plc2_employees pe2 WHERE pe2.Department = pe1.Department)
 CONNECT BY PRIOR pe1.Department = pe1.Department
        AND PRIOR pe1.Salary > pe1.Salary
        AND PRIOR pe1.Salary = ( SELECT MIN(Salary) FROM plc2_employees pe3
                                  WHERE pe3.Department = pe1.Department
                                    AND pe3.Salary > pe1.Salary
                               )

递归条件表明子行和父行之间没有中间工资。

请注意，这确实是低效的……

Answer 4

试试这个，它给了第二高的薪水

select MAX(Salary) as Salary 
from Employee_salary 
where Salary not in (select MAX(Salary) from Employee_salary) 

Answer 5

你可以使用这个查询：

select * from 
employee e1 
where 2 = (select count (distinct (salary)) 
from employee e2 
where e2.salary >=e1.salary);

Answer 6

从以salary为列的employee表中找出第二高的薪水： 数据库：DB2

with t as
(
select distinct salary from employee order by salary desc
),
tr as 
(
select salary, row_Number() over() r from t 
)
select salary from tr where r = 2

Answer 7

试试这个， 它给了第二高的薪水......

select MAX(Salary) as Salary 
from Employee_salary 
where Salary not in (select MAX(Salary) from Employee_salary )

如果你想找到第 n 高的薪水，你可以使用以下查询.... 你只需要做一个改变...... 把 N=n 的值放最高

干杯....:)

SELECT * FROM Employee_salary Emp1 
WHERE (N-1) = (SELECT COUNT(DISTINCT(Emp2.Salary)) 
               FROM Employee_salary Emp2 
               WHERE Emp2.Salary > Emp1.Salary)

Answer 8

这会起作用 -

SELECT MIN(Salary) 
FROM employee 
WHERE salary IN (SELECT TOP 2 salary FROM employee ORDER BY salary DESC)

Answer 9

首先，按降序（从大到小）选择不同的薪水，从该集合中选择前 2 个并按升序排列（将数字 2 放在顶部），然后从这两个中选择前 1 个：

select top 1 s.Salary
  from 
   (select top 2 t.Salary
      from
          (select distinct Salary
             from PLC2_Employees
            order by Salary desc) t
     order by Salary asc) s