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)