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
AND C.MgrID=P.EmpID
END
RETURN;
END
SELECT * FROM fn_GetEmployeeTree(1)
本文转自baoqiangwang51CTO博客，原文链接：http://blog.51cto.com/baoqiangwang/310454，如需转载请自行联系原作者
