关于SQLServer2005的学习笔记――树形结构问题

    关于解决树形目录是每种数据库 或大多数开发人员都要面对的问题，在这一点上 Oracle 走的跟前线一些，从 9i 起便提供了 connect by 进行了支持， 10g 又增强了相关语法；在 sqlServer2005 中，强大的 CTE 功能也提供了相应的解决方案，此外提供的表函数功能也给出了另外一种解决思路。

 

从功能上讲的话，表函数方式更为灵活一些，毕竟基于过程的结构方式更容易实现负责的业务逻辑；但递归 CTE 构造起来更为清晰一些。

 

该文起源于《 Microsoft sql Server 2005 技术内幕： T-sql 查询》，但与文中所述不尽相同。

首先构建一个标准的树形结构的员工表

CREATE TABLE Employees

(

  EmpID        INT,

  MgrID        INT,

  EmpName      VARCHAR(25),

  Salary       MONEY,

  CHECK(EmpID<>MgrID)

);

GO

INSERT INTO Employees VALUES(1,NULL,'David',10000);

INSERT INTO Employees VALUES(2,1,'Eitan',7000);

INSERT INTO Employees VALUES(3,'Ina',7500);

INSERT INTO Employees VALUES(4,2,'Seraph',5000);

INSERT INTO Employees VALUES(5,'Jiru',5500);

INSERT INTO Employees VALUES(6,'Steve',4500);

INSERT INTO Employees VALUES(7,3,'Aaron',5000);

INSERT INTO Employees VALUES(8,5,'Lilach',3500);

INSERT INTO Employees VALUES(9,7,'Rita',3000);

INSERT INTO Employees VALUES(10,'Sean',3000);

INSERT INTO Employees VALUES(11,'Gabriel',3000);

INSERT INTO Employees VALUES(12,9,'Emilia',2000);

INSERT INTO Employees VALUES(13,'Michael',2000);

INSERT INTO Employees VALUES(14,'Didi',1500);
 

 

-- 让我们先来看看 Oracle 是如何实现的吧

-- 获取所有相关员工信息，并构建其级别和相应的结构指向

SELECT EmpID,MgrID,EmpName,Salary,Level,sys_connect_by_path(NVL(EmpID,'0'),'->')

  FROM Employees

CONNECT BY PRIOR EmpID=MgrID

  START WITH MgrID IS NULL

-- 获取员工的所有下级节点

SELECT EmpID,Level

  FROM Employees

CONNECT BY PRIOR EmpID=MgrID

  START WITH EmpID=9

-- 获取员工的所有上级节点

SELECT EmpID,Level

  FROM Employees

CONNECT BY PRIOR MgrID=EmpID

  START WITH EmpID=14
 

 

-- 构建递归 CTE ，也可以灵活获取满足不同级别的上下级节点

WITH EmployeeTree

AS

(

  SELECT EmpID,

         0 AS Level,

         CAST(CASE WHEN MgrID IS NULL THEN 'Root' END AS VARCHAR(50)) MgrList

    FROM Employees

   WHERE MgrID IS NULL -- 此处亦可修改为 MgrID=@Root ，即传入的节点，即可得到想要的节点内容

  UNION ALL

  SELECT C.EmpID,C.MgrID,C.EmpName,C.Salary,

         P.Level+1 AS Level,

         CAST(CAST(P.MgrList AS VARCHAR(50))+'->'+CAST(C.EmpID AS VARCHAR(10)) AS VARCHAR(50)) MgrList

   FROM EmployeeTree P,Employees C

   WHERE C.MgrID=P.EmpID --AND P.Level<2 设定相关级别

)

-- 所有员工

SELECT * FROM EmployeeTree

-- 求某员工上级

SELECT * FROM EmployeeTree

  WHERE CHARINDEX(MgrList,(SELECT MgrList FROM EmployeeTree WHERE EmpID=7))>0

-- 求某员工下级

SELECT * FROM EmployeeTree

  WHERE MgrList LIKE (SELECT MgrList FROM EmployeeTree WHERE EmpID=7)+'%'

-- 求某员工下级并且符合相应级数的

SELECT * FROM EmployeeTree

  WHERE MgrList LIKE (SELECT MgrList FROM EmployeeTree WHERE EmpID=7)+'%'

   AND Level<=(SELECT Level FROM EmployeeTree WHERE EmpID=7)+1
 

 

-- 通过表函数方式返回相关节点

CREATE FUNCTION fn_GetEmployeeTree(@root AS INT)

RETURNS @Subs TABLE

(

  EmpID INT,

  Level INT

)

AS

BEGIN

  DECLARE @Level AS INT;

  SET @Level=0;

  INSERT INTO @Subs(EmpID,Level) SELECT EmpID,@Level FROM Employees WHERE EmpID=@root;

 

  WHILE @@rowcount>0

  BEGIN

    SET @Level=@Level+1;

    INSERT INTO @Subs(EmpID,Level)

    SELECT C.EmpID,@Level

      FROM @SubS AS P

      JOIN Employees AS C

        ON P.Level=@Level-1

       AND C.MgrID=P.EmpID

  END

   

  RETURN;

 

END

SELECT * FROM fn_GetEmployeeTree(1)