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(3,1,'Ina',7500); INSERT INTO Employees VALUES(4,2,'Seraph',5000); INSERT INTO Employees VALUES(5,2,'Jiru',5500); INSERT INTO Employees VALUES(6,2,'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(10,5,'Sean',3000); INSERT INTO Employees VALUES(11,7,'Gabriel',3000); INSERT INTO Employees VALUES(12,9,'Emilia',2000); INSERT INTO Employees VALUES(13,9,'Michael',2000); INSERT INTO Employees VALUES(14,9,'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,MgrID,EmpName,Salary,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,MgrID,EmpName,Salary,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,MgrID,EmpName,Salary,          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)