在RDBMS中操作 圖 樹 層次結構 等特殊的資料結構時,我們通常採用2個主要方法:
1.基於迭代/遞迴

2.具體化描述資料結構的附加資訊。
一般模型有：員工組織圖(樹,層次結構);料表--BOM(有向圖);道路系統(無向迴圈圖)

1.迭代/遞迴
迭代可以迭代圖的一個節點,也可以迭代一個層次.後者比前者要快很多.
實現方法：SQL2000通過UDF(使用者自定義函式),SQL2005使用CTE。

a.下屬問題(通俗說,求子節點)
--這裡我使用書上的員工表(表內容如下)
SET NOCOUNT ON; USE tempdb; GO IF OBJECT_ID('dbo.Employees') IS NOT NULL DROP TABLE dbo.Employees; GO CREATE TABLE dbo.Employees ( empid INT NOT NULL PRIMARY KEY, mgrid INT NULL REFERENCES dbo.Employees, empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, CHECK (empid <> mgrid) ); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(1, NULL, 'David', $10000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(2, 1, 'Eitan', $7000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(3, 1, 'Ina', $7500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(4, 2, 'Seraph', $5000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(5, 2, 'Jiru', $5500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(6, 2, 'Steve', $4500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(7, 3, 'Aaron', $5000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(8, 5, 'Lilach', $3500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(9, 7, 'Rita', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(10, 5, 'Sean', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(11, 7, 'Gabriel', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(12, 9, 'Emilia' , $2000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(13, 9, 'Michael', $2000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(14, 9, 'Didi', $1500.00); --建立索引 CREATE UNIQUE INDEX idx_unc_mgrid_empid ON dbo.Employees(mgrid, empid); go

--SQL2000 udf方法：

IF OBJECT_ID('dbo.fn_subordinates1') IS NOT NULL DROP FUNCTION dbo.fn_subordinates1; GO CREATE FUNCTION dbo.fn_subordinates1(@root AS INT) RETURNS @Subs TABLE ( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) ) AS begin declare @lv int set @lv=0 insert @Subs values(@root,@lv) while @@rowcount>0 begin set @lv=@Lv+1; insert @subs select b.empid ,@Lv from @subs a join dbo.Employees b on a.empid=b.mgrid and lvl=@lv-1 end return; end go SELECT empid, lvl FROM dbo.fn_subordinates1(3) AS S;

--SQL2005 CTE

DECLARE @root AS INT; SET @root = 3; WITH SubsCTE AS ( -- Anchor member returns root node SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL -- Recursive member returns next level of children SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid ) SELECT * FROM SubsCTE;

--查詢結果
/* empid empname lvl ----------- ------------------------- ----------- 3 Ina 0 7 Aaron 1 9 Rita 2 11 Gabriel 2 12 Emilia 3 13 Michael 3 14 Didi 3 */

---------------如果需要限制遞迴的層數------------

--SQL2000

IF OBJECT_ID('dbo.fn_subordinates2') IS NOT NULL DROP FUNCTION dbo.fn_subordinates2; GO CREATE FUNCTION dbo.fn_subordinates2 (@root AS INT, @maxlevels AS INT = NULL) RETURNS @Subs TABLE ( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) ) AS BEGIN DECLARE @lvl AS INT; SET @lvl = 0; -- 如果@maxlevels輸入的是NULL,把他變為最大的INT型別 SET @maxlevels = COALESCE(@maxlevels, 2147483647); INSERT INTO @Subs(empid, lvl) SELECT empid, @lvl FROM dbo.Employees WHERE empid = @root; WHILE @@rowcount > 0 AND @lvl < @maxlevels -- 這裡注意加一個小於最大的遞迴數的條件,是小於 不是小於等於 BEGIN SET @lvl = @lvl + 1; INSERT INTO @Subs(empid, lvl) SELECT C.empid, @lvl --這裡跟上面處理都一樣 FROM @Subs AS P JOIN dbo.Employees AS C ON P.lvl = @lvl - 1 AND C.mgrid = P.empid; END RETURN; END GO SELECT empid, lvl FROM dbo.fn_subordinates2(3, NULL) AS S;

--查詢結果

/* empid lvl ----------- ----------- 3 0 7 1 9 2 11 2 12 3 13 3 14 3 */ -----

SELECT empid, lvl
FROM dbo.fn_subordinates2(2, NULL) AS S;
/* empid lvl ----------- ----------- 2 0 4 1 5 1 6 1 8 2 10 2 */

PS：這裡控制返回的層數 你當然可以用最開始的方法,然後再篩選語句裡控制層數
 如SELECT empid, lvl FROM dbo.fn_subordinates1(3) AS S where lvl<3;  但是控制本身的遞迴只能在UDF裡面了
 
 --sql2005方法類似

WITH SubsCTE AS ( SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid AND P.lvl < @maxlevels -- 這裡控制遞迴數 ) SELECT * FROM SubsCTE;

--還有一種偏方：但是不推薦 雖然結果正確 但是會報錯
WITH SubsCTE AS ( SELECT empid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1 FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid ) SELECT * FROM SubsCTE OPTION (MAXRECURSION 2);--就是這裡的MAXRECURSION 控制遞迴

--查詢結果

/* empid empname lvl ----------- ------------------------- ----------- 3 Ina 0 7 Aaron 1 9 Rita 2 11 Gabriel 2 訊息 530，級別 16，狀態 1，第 4 行 語句被終止。完成執行語句前已用完最大遞迴 2。 */ 
 

b.祖先(通俗說：求父節點)
其實思路跟子節點差不多

--SQL2000

IF OBJECT_ID('dbo.fn_managers') IS NOT NULL DROP FUNCTION dbo.fn_managers; GO CREATE FUNCTION dbo.fn_managers (@empid AS INT, @maxlevels AS INT = NULL) RETURNS @Mgrs TABLE ( empid INT NOT NULL PRIMARY KEY, lvl INT NOT NULL ) AS BEGIN IF NOT EXISTS(SELECT * FROM dbo.Employees WHERE empid = @empid) --這裡判斷是否存在經理,不存在 馬上返回 RETURN; DECLARE @lvl AS INT,@mgrid int ; SET @lvl = 0; set @mgrid=@empid --賦值給@mgrid 我這是為了更加清楚變數的意思 其實不用這個@Mgrid變數的 -- 如果@maxlevels輸入的是NULL,把他變為最大的INT型別 SET @maxlevels = COALESCE(@maxlevels, 2147483647); WHILE @lvl <= @maxlevels and @mgrid is not null --注意這裡是小於等於 而且新插入的經理不可以為NULL，為null說明是老大了 BEGIN --插入當前經理 INSERT INTO @Mgrs(empid, lvl) VALUES(@mgrid, @lvl); --這裡第一次插入是自己,後面就是各自的經理了 SET @lvl = @lvl + 1; --獲得下一個級別的經理 SET @mgrid = (SELECT mgrid FROM dbo.Employees WHERE empid = @mgrid); END RETURN; END GO SELECT empid, lvl FROM dbo.fn_managers(8, NULL) AS M;

--查詢結果

/* empid lvl ----------- ----------- 1 3 2 2 5 1 8 0 */

SELECT empid, lvl
FROM dbo.fn_managers(8, 2) AS M;
/* empid lvl ----------- ----------- 2 2 5 1 8 0 */

--sql2005
DECLARE @empid AS INT, @maxlevels AS INT; SET @empid = 8; SET @maxlevels = 2; WITH MgrsCTE AS ( SELECT empid, mgrid, empname, 0 AS lvl FROM dbo.Employees WHERE empid = @empid UNION ALL SELECT P.empid, P.mgrid, P.empname, C.lvl + 1 FROM MgrsCTE AS C JOIN dbo.Employees AS P ON C.mgrid = P.empid AND C.lvl < @maxlevels--限制遞迴數... ) SELECT * FROM MgrsCTE;

C.層次顯示

--SQL2000,思路跟下屬問題一模一樣 只是多了個PATH
IF OBJECT_ID('dbo.fn_subordinates3') IS NOT NULL DROP FUNCTION dbo.fn_subordinates3; GO CREATE FUNCTION dbo.fn_subordinates3 (@root AS INT, @maxlevels AS INT = NULL) RETURNS @Subs TABLE ( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, lvl INT NOT NULL, path VARCHAR(900) NOT NULL UNIQUE CLUSTERED(lvl, empid) ) AS BEGIN DECLARE @lvl AS INT; SET @lvl = 0; SET @maxlevels = COALESCE(@maxlevels, 2147483647); INSERT INTO @Subs(empid, lvl, path) SELECT empid, @lvl, CAST(empid AS VARCHAR(100))--ZHU YI ZHE LI FROM dbo.Employees WHERE empid = @root; WHILE @@rowcount > 0 AND @lvl < @maxlevels BEGIN SET @lvl = @lvl + 1; INSERT INTO @Subs(empid, lvl, path) SELECT C.empid, @lvl, P.path +'-'+ CAST(C.empid AS VARCHAR(100)) --和上面型別保持一致 FROM @Subs AS P JOIN dbo.Employees AS C ON P.lvl = @lvl - 1 AND C.mgrid = P.empid; END RETURN; END GO

--這裡的顯示分2種
--顯示1
select empid ,pos=REPLICATE('-',lvl)+rtrim(empid)
FROM dbo.fn_subordinates3(1, NULL) AS S
ORDER BY PATH
/* empid pos ----------- ------------ 1 1 2 -2 4 --4 5 --5 10 ---10 8 ---8 6 --6 3 -3 7 --7 11 ---11 9 ---9 12 ----12 13 ----13 14 ----14 */

--還有一種
SELECT empid,  path
FROM dbo.fn_subordinates3(1, NULL) AS S
ORDER BY PATH
/* empid path ----------- ---------------- 1 1 2 1-2 4 1-2-4 5 1-2-5 10 1-2-5-10 8 1-2-5-8 6 1-2-6 3 1-3 7 1-3-7 11 1-3-7-11 9 1-3-7-9 12 1-3-7-9-12 13 1-3-7-9-13 14 1-3-7-9-14 */

--當然上面的顯示方式還很多,自己可以控制 比如不加ORDER 就可以排出不一樣的

--SQL2005
DECLARE @root AS INT; SET @root = 1; WITH SubsCTE AS ( SELECT empid, empname, 0 AS lvl, -- Path of root = '.' + empid + '.' CAST(CAST(empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1, CAST(P.path+'-' + CAST(C.empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid ) SELECT empid, REPLICATE(' | ', lvl) + empname AS empname FROM SubsCTE ORDER BY path ;

--結果查詢

/* empid empname ----------- ------------------------- 1 David 2 | Eitan 4 | | Seraph 5 | | Jiru 10 | | | Sean 8 | | | Lilach 6 | | Steve 3 | Ina 7 | | Aaron 11 | | | Gabriel 9 | | | Rita 12 | | | | Emilia 13 | | | | Michael 14 | | | | Didi */

D.檢測迴圈中異常
說白了 就是檢查表裡有沒出現1-2-4-1這種頭接尾的圈.這在現實中是不可能的.一個經理部可能是它手下的手下.-- ||
我們對錶做手腳,把老大的經理改成是它的一個員工
UPDATE dbo.Employees SET mgrid = 14 WHERE empid = 1; DECLARE @root AS INT; SET @root = 1; WITH SubsCTE AS ( SELECT empid, empname, 0 AS lvl, CAST( CAST(empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path, -- 第一層是不會出現圈圈的 0 AS cycle --0表示沒有圈 1表示出現圈 FROM dbo.Employees WHERE empid = @root UNION ALL SELECT C.empid, C.empname, P.lvl + 1, CAST(P.path + '-'+CAST(C.empid AS VARCHAR(10)) AS VARCHAR(MAX)) AS path, -- 這裡需要檢查 CASE WHEN P.path LIKE '%' + CAST(C.empid AS VARCHAR(10)) + '%' THEN 1 ELSE 0 END FROM SubsCTE AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid where p.cycle=0 --保證不會繼續在錯誤的地方繼續迴圈下去 ) SELECT empid, empname, cycle, path FROM SubsCTE where cycle=1

--查詢結果

/* empid empname cycle path ----------- ------------------------- ----------- ------------------ 1 David 1 1-3-7-9-14-1 */
--這樣管理員可以輕易找到那個錯誤的地方.
改過來：UPDATE dbo.Employees SET mgrid = NULL WHERE empid = 1;

2.具體化路徑
這裡就是新增加2列，一列是級別 一列是節點路徑，這樣可以避免每次都去計算.
2個優點：不需要遞迴,只需要基於集合 ；查詢可以使用到路徑索引

a.維護資料

1.新增不管理員工的員工
SET NOCOUNT ON; USE tempdb; GO IF OBJECT_ID('dbo.Employees') IS NOT NULL DROP TABLE dbo.Employees; GO CREATE TABLE dbo.Employees ( empid INT NOT NULL PRIMARY KEY NONCLUSTERED, mgrid INT NULL REFERENCES dbo.Employees, empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, lvl INT NOT NULL, path VARCHAR(900) NOT NULL UNIQUE CLUSTERED ); CREATE UNIQUE INDEX idx_unc_mgrid_empid ON dbo.Employees(mgrid, empid); GO IF OBJECT_ID('dbo.usp_insertemp') IS NOT NULL DROP PROC dbo.usp_insertemp; GO CREATE PROC dbo.usp_insertemp @empid INT, @mgrid INT, @empname VARCHAR(25), @salary MONEY AS SET NOCOUNT ON; -- 如果插入的是一個沒有經理的老大 IF @mgrid IS NULL INSERT INTO dbo.Employees(empid, mgrid, empname, salary, lvl, path) VALUES(@empid, @mgrid, @empname, @salary, 0, CAST(@empid AS VARCHAR(10))); --插入有經理的小弟 ELSE INSERT INTO dbo.Employees(empid, mgrid, empname, salary, lvl, path) SELECT @empid, @mgrid, @empname, @salary, lvl + 1, path +'-'+ CAST(@empid AS VARCHAR(10)) FROM dbo.Employees WHERE empid = @mgrid; GO EXEC dbo.usp_insertemp @empid = 1, @mgrid = NULL, @empname = 'David', @salary = $10000.00; EXEC dbo.usp_insertemp @empid = 2, @mgrid = 1, @empname = 'Eitan', @salary = $7000.00; EXEC dbo.usp_insertemp @empid = 3, @mgrid = 1, @empname = 'Ina', @salary = $7500.00; EXEC dbo.usp_insertemp @empid = 4, @mgrid = 2, @empname = 'Seraph', @salary = $5000.00; EXEC dbo.usp_insertemp @empid = 5, @mgrid = 2, @empname = 'Jiru', @salary = $5500.00; EXEC dbo.usp_insertemp @empid = 6, @mgrid = 2, @empname = 'Steve', @salary = $4500.00; EXEC dbo.usp_insertemp @empid = 7, @mgrid = 3, @empname = 'Aaron', @salary = $5000.00; EXEC dbo.usp_insertemp @empid = 8, @mgrid = 5, @empname = 'Lilach', @salary = $3500.00; EXEC dbo.usp_insertemp @empid = 9, @mgrid = 7, @empname = 'Rita', @salary = $3000.00; EXEC dbo.usp_insertemp @empid = 10, @mgrid = 5, @empname = 'Sean', @salary = $3000.00; EXEC dbo.usp_insertemp @empid = 11, @mgrid = 7, @empname = 'Gabriel', @salary = $3000.00; EXEC dbo.usp_insertemp @empid = 12, @mgrid = 9, @empname = 'Emilia', @salary = $2000.00; EXEC dbo.usp_insertemp @empid = 13, @mgrid = 9, @empname = 'Michael', @salary = $2000.00; EXEC dbo.usp_insertemp @empid = 14, @mgrid = 9, @empname = 'Didi', @salary = $1500.00 ;
----檢測
SELECT empid, mgrid, empname, salary, lvl, path
FROM dbo.Employees
ORDER BY path;
/* empid mgrid empname salary lvl path ----------- ----------- ------------------------- --------------------- ----------- --------- 1 NULL David 10000.00 0 1 2 1 Eitan 7000.00 1 1-2 4 2 Seraph 5000.00 2 1-2-4 5 2 Jiru 5500.00 2 1-2-5 10 5 Sean 3000.00 3 1-2-5-10 8 5 Lilach 3500.00 3 1-2-5-8 6 2 Steve 4500.00 2 1-2-6 3 1 Ina 7500.00 1 1-3 7 3 Aaron 5000.00 2 1-3-7 11 7 Gabriel 3000.00 3 1-3-7-11 9 7 Rita 3000.00 3 1-3-7-9 12 9 Emilia 2000.00 4 1-3-7-9-12 13 9 Michael 2000.00 4 1-3-7-9-13 14 9 Didi 1500.00 4 1-3-7-9-14 */

2.移動子樹
比如說某個部門來了個新老大,原來部門老大和手下的人都要跟著他.這個時候表裡的路徑和級別都要更新
Select Empid, Replicate(' | ', Lvl) + Empname As Empname, Lvl, Path
From Dbo.Employees
Order By Path;

--這個是移動之前的層次分佈

/* empid empname lvl path ----------- ----------------- ----------- ------------ 1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 7 | | Aaron 2 1-3-7 11 | | | Gabriel 3 1-3-7-11 9 | | | Rita 3 1-3-7-9 13 | | | | Michael 4 1-3-7-9-13 14 | | | | Didi 4 1-3-7-9-14 */

==接下來我們移動
IF OBJECT_ID('dbo.usp_movesubtree') IS NOT NULL DROP PROC dbo.usp_movesubtree; GO CREATE PROC dbo.usp_movesubtree @root INT,--舊經理 @mgrid INT--新經理 AS SET NOCOUNT ON; BEGIN TRAN; UPDATE E SET lvl = E.lvl + NM.lvl - OM.lvl,-- 級別=當前的級別+(新經理級別-原經理級別) path = STUFF(E.path, 1, LEN(OM.path), NM.path) -- 路徑=自己的路徑移除原來經理那部分再加上新的經理的那部分 --這裡的 OM.path 是被替換經理的經理的路徑 比如舊經理是1-3-7 那麼OM.PATH 就是1-3 --NM.path 是新經理的路徑了 比如要換新經理的EMPID是10 所以NM.path 就是 1-2-5-10 --所以如果本來舊經理一個手下比如說是1-3-7-10 現在整個替換過來就是1-2-5-10-7-10 FROM dbo.Employees AS E -- E = 員工 JOIN dbo.Employees AS R -- R = 根 ON R.empid = @root AND E.path LIKE R.path + '%' JOIN dbo.Employees AS OM -- OM = 舊經理 ON OM.empid = R.mgrid JOIN dbo.Employees AS NM -- NM = 新經理 ON NM.empid = @mgrid; -- 更新舊的經理的經理為新來的經理 UPDATE dbo.Employees SET mgrid = @mgrid WHERE empid = @root; COMMIT TRAN; GO BEGIN TRAN; EXEC dbo.usp_movesubtree @root = 7, @mgrid = 10; -- After moving subtree SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, path FROM dbo.Employees ORDER BY path; ROLLBACK TRAN;

--移動後結果

/* empid empname lvl path ----------- ------------------------------ ----------- ---------------------- 1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 7 | | | | Aaron 4 1-2-5-10-7 11 | | | | | Gabriel 5 1-2-5-10-7-11 9 | | | | | Rita 5 1-2-5-10-7-9 12 | | | | | | Emilia 6 1-2-5-10-7-9-12 13 | | | | | | Michael 6 1-2-5-10-7-9-13 14 | | | | | | Didi 6 1-2-5-10-7-9-14 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 */

--這個是移動之後 大家注意觀察7 和 10 兩位同志

3.移除子樹
就是把某個部門從公司取消掉
--首先檢視公司部門當前分佈
SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, path FROM dbo.Employees ORDER BY path;

--查詢結果
/* empid empname lvl path ----------- ------------- ------------- ----------- 1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 7 | | Aaron 2 1-3-7 11 | | | Gabriel 3 1-3-7-11 9 | | | Rita 3 1-3-7-9 12 | | | | Emilia 4 1-3-7-9-12 13 | | | | Michael 4 1-3-7-9-13 14 | | | | Didi 4 1-3-7-9-14 */
--接著我們來開始一個部門 比如移除Aaron手下的人

BEGIN TRAN; DELETE FROM dbo.Employees WHERE path LIKE (SELECT M.path + '%' FROM dbo.Employees as M WHERE M.empid = 7); --刪除之後顯示 SELECT empid, REPLICATE(' | ', lvl) + empname AS empname, lvl, path FROM dbo.Employees ORDER BY path; ROLLBACK TRAN;

--查詢結果

/* empid empname lvl path ----------- ------------------ ----------- ------------ 1 David 0 1 2 | Eitan 1 1-2 4 | | Seraph 2 1-2-4 5 | | Jiru 2 1-2-5 10 | | | Sean 3 1-2-5-10 8 | | | Lilach 3 1-2-5-8 6 | | Steve 2 1-2-6 3 | Ina 1 1-3 */

4.查詢

這裡的查詢可就輕鬆多啦 ~因為路徑都有啦~
--查詢EMPID為3的手下一批人

SELECT REPLICATE(' | ', E.lvl - M.lvl) + E.empname FROM dbo.Employees AS E JOIN dbo.Employees AS M ON M.empid = 3 -- root AND E.path LIKE M.path + '%' --and E.path LIKE M.path + '_%'--這裡那個3號就沒了 --and E.lvl - M.lvl <= 2 --限制級數 --WHERE NOT EXISTS --自己本身不是經理的,返回的都是員工 (SELECT * FROM dbo.Employees AS E2 WHERE E2.mgrid = E.empid); ORDER BY E.path;

--查詢結果
/* ------------------ Ina | Aaron | | Gabriel | | Rita | | | Emilia | | | Michael | | | Didi */

b.巢狀集合
是書的作者認為用於樹建模最完美最高明的解決方案.
慚愧..太難..準備翻第二次書的時候再研究...