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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
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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)
（编辑：李大同）
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