Question

我有下表

empid   empname     managerID
1          A          NULL
2          B           1
3          C           1
4          D           2

ddl在

之下

Declare @t table(empid int, empname varchar(20), managerID int)
insert into @t 

select 1,'A',null union all
select 2,'B',1 union all
select 3,'C',1 union all 
select 4,'D',2

我要做的是准备一份报告，说明哪些员工向哪位经理报告。

我已经使用

进行了宣传

select EmpName = e.empname, ManagerName = m.empname
from @t e 
left join
@t m
on  e.managerID = m.empid

，所需的输出为

EmpName    ManagerName
A          NULL
B          A
C          A
D          B

这样做的其他方法是什么？

Answer 1

Declare @t table(empid int, empname varchar(20), managerID int)

insert into @t 
select 1,'A',null union all
select 2,'B',1 union all
select 3,'C',1 union all 
select 4,'D',2

;with CTE AS
(
    Select empid,empname,managerID,
    Cast(empname As Varchar(max)) As ReportingManager
    from @T
    where managerID is null

    UNION ALL

    Select T.empid,T.empname,T.managerID,
    Cast(CTE.empname+'->'+t.empname As Varchar(max)) As ReportingManager
    from @T As T
    INNER JOIN CTE ON T.managerID=CTE.empid 
)
SELECT *
FROM CTE

Answer 2

我认为，您的解决方案是最合适的，但我们可以重写它，例如：

select t1.empname [EmpName]
    , (select t2.empname from @t t2 where t2.empid = t1.managerID) [ManagerName]
from @t t1

Answer 3

自我加入表格

案例1：所有拥有经理的员工

select e1.empname employee, e2.empname manager
from employee e1
join employee e2 on e1.managerId = e2.empid
order by e2.name,e1.name

案例2：所有拥有经理的员工

select e1.empname employee, COALESCE(e2.empname,'none') manager
from employee e1
left join employee e2 on e1.managerId = e2.empid
order by e2.name,e1.name

Answer 4

好的，所以你问其他方法。这有点怪异。

CREATE TABLE employee (empid int, empname varchar(20), managerID int)
GO
insert into employee 

select 1,'A',null union all
select 2,'B',1 union all
select 3,'C',1 union all 
select 4,'D',2
GO

CREATE FUNCTION [dbo].[GetEmployeeTree](@ManagerId int)
RETURNS XML
WITH RETURNS NULL ON NULL INPUT 
BEGIN RETURN 
  (SELECT   empID as '@Id', 
            empname AS '@Name',
            dbo.GetEmployeeTree(empid) 
   FROM employee em
   WHERE ManagerId=@ManagerId
   FOR XML PATH('Employee'), TYPE)
END

GO
SELECT  empID as '@Id', 
        empname AS '@Name',
        dbo.GetEmployeeTree(empId) 
FROM employee
WHERE managerId is null
FOR XML PATH('Employee'), ROOT('Employees')

提供此输出

<Employees>
  <Employee Id="1" Name="A">
    <Employee Id="2" Name="B">
      <Employee Id="4" Name="D" />
    </Employee>
    <Employee Id="3" Name="C" />
  </Employee>
</Employees>

我实际上已经使用它来生成具有数万个节点的大型XML树，而且速度非常快。可能有一种方法将根查询与子查询合并，我还没有完全弄明白。当我过去使用这种技术时，我使用了一个单独的链接和节点表来定义层次结构，当你这样做时它会更清晰。

员工经理报告 - Sql Server

4 个答案: