Question

我需要获取一个查询，从一个特定的经理开始一直显示下属。让我们说我有这种结构，我需要让所有员工从ManagerId = 1

开始

EmployeeId  ManagerId
2           1
3           1 
4           3
5           3 
6           4

我需要获得ManagerId的查询结果，并直接向他下属的员工及其下属报告。

例如，使用SQL Server 2014：

CREATE TABLE #Pepe (EmployeeId INT, ManagerId int)

INSERT INTO [#Pepe] ( [EmployeeId], [ManagerId] ) VALUES  (2, 1)
INSERT INTO [#Pepe] ( [EmployeeId], [ManagerId] ) VALUES  (3, 1)
INSERT INTO [#Pepe] ( [EmployeeId], [ManagerId] ) VALUES  (4, 3)
INSERT INTO [#Pepe] ( [EmployeeId], [ManagerId] ) VALUES  (5, 3)
INSERT INTO [#Pepe] ( [EmployeeId], [ManagerId] ) VALUES  (6, 4)

现在我得到了CTE

;WITH relation AS 
    (
       SELECT 1 as EmployeeId,  0 AS LEVEL

       UNION ALL

       SELECT r.EmployeeId, LEVEL + 1 AS LEVEL
         FROM (SELECT EmployeeId, ManagerId FROM #Pepe) r
        INNER JOIN relation T 
           ON r.ManagerId = T.EmployeeId
        WHERE r.ManagerId <> r.EmployeeId 

    )  

SELECT DISTINCT  EmployeeId, LEVEL FROM relation

我的CTE结果如下：

EmployeeId  LEVEL
1           0
2           1 
3           1
4           2
5           2
6           3

此结果从一个特定员工开始，该员工为1（在CTE上是硬编码的），现在我只需要直接报告，Level = 1以及也是0级员工ID 1,2和3。

这很好，现在我需要的是每个员工ID（0级和1级），我需要得到这个结果：

EmployeeId  ManagerId
1           1
2           1
3           1
4           1
5           1
6           1
2           2
3           3
4           3
5           3
6           3

正如您所看到的，Manager ID包含0级和1级的员工，对于每个员工，我基本上都会调用CTE来让所有员工失望，例如ManagerId = 2没有下属，但我需要无论如何要数。有没有一种有效的方法来做到这一点？我正在使用交叉应用将CTE放入内联函数中，但我遇到了性能问题。

Answer 1

我想我找到了解决方案，我想就此发表意见。

;WITH relation AS 
    (
       SELECT 1 AS ManagerId, 1 as EmployeeId,  0 AS LEVEL UNION    
       SELECT 2 AS ManagerId, 2 as EmployeeId,  0 AS LEVEL UNION
       SELECT 3 AS ManagerId, 3 as EmployeeId,  0 AS LEVEL

       UNION ALL

       SELECT t.[ManagerId], r.EmployeeId, LEVEL + 1 AS LEVEL
         FROM (SELECT EmployeeId, ManagerId FROM #Pepe) r
        INNER JOIN relation T 
           ON r.ManagerId = T.EmployeeId
        WHERE r.ManagerId <> r.EmployeeId 

    )  

SELECT DISTINCT EmployeeId, [ManagerId] FROM relation ORDER BY 2

Answer 2

您几乎拥有了它。只需从#Pepe表中选择不同的经理，而不是手动执行3个select语句

DECLARE @Pepe TABLE (EmployeeId INT, ManagerId int)

INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (1, 1)
INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (2, 1)
INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (3, 1)
INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (4, 3)
INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (5, 3)
INSERT INTO @Pepe ( [EmployeeId], [ManagerId] ) VALUES  (6, 4)

;WITH relation (ManagerId, EmployeeId, LEVEL)
AS 
(
    SELECT DISTINCT ManagerId, ManagerId AS EmployeeId,  0 AS LEVEL
    FROM @Pepe  

    UNION ALL

    SELECT t.[ManagerId], r.EmployeeId, LEVEL + 1 AS LEVEL
    FROM @Pepe r
      INNER JOIN relation T 
        ON r.ManagerId = T.EmployeeId
    WHERE r.ManagerId <> r.EmployeeId 
)  
SELECT DISTINCT EmployeeId, ManagerId, LEVEL FROM relation ORDER BY ManagerId

如何查找每位经理下的所有员工

2 个答案: