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薪

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

某种程度上不起作用......我没有得到正确的结果。有谁可以帮帮我。

Answer 1

像

这样的东西

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 - 测试它，它给出了你期望的答案。

Answer 2

如果您要自学如何处理CONNECT BY，您应该首先找到适合构造的问题。 CONNECT BY用于处理分层形式的数据，而您的示例则不是。工资以分层方式彼此无关。试图在错误的问题上强制构建一个构造是令人沮丧的，并没有真正教你什么。

查看可以与Oracle一起安装的演示HR模式中的经典员工 - 经理关系。所有员工都向经理报告，包括经理（除了顶级人员）。然后，您可以使用此架构创建查询，以显示公司的组织结构图。

Answer 3

START WITH … CONNECT BY旨在通过探索所有可能的降序路径来探索形成图表的数据。您可以在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子句过滤器将在层次条件后处理，GROUP BY和HAVING相同（当然因为计算GROUP BY在WHERE）之后。

因此，你必须在这里CONNECT BY PRIOR department = department。当存在中间工资时，您还必须避免在两个工资之间完成节点连接。

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，这实际上是无效的......

Answer 4

试试这个，它给出了第二高的薪水

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

Answer 5

试试这个，它的薪水排名第二......

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个最高薪水，你可以使用以下查询.... 你需要做一个改变..... 将N = n的值设为最高

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 6

从员工表中找到第二高的薪水，列为薪水： 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

这可行 -

从员工中选择最低（薪水）

（按工资描述从员工订单中选择前2名薪水）

Answer 8

您可以使用此查询：

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

Answer 9

首先，按降序选择不同的工资（从最大到最小），从该组中选择前2并按升序排列（将数字2放在顶部），然后从这2个中选择前1：

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

找到第二高薪

9 个答案: