您好我在尝试制作一个视图，从单个表中保存（黑手党）层次结构（最多5个级别），每个用户都有一个额外的表来填写每个用户的名称，这就是实际结构，我使用的是windows azure。
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这些是我从以下网站获取数据的表格：
dbo.userFamily dbo.user
id | user_id | parent_id id_user | name
------------------------ ------------------
1 | 2 | 1 1 | Victor
2 | 3 | 1 2 | Lucifer
3 | 4 | 1 3 | Hellboy
4 | 11 | 2 4 | Spiderman
5 | 12 | 2 5 | Martin
6 | 10 | 4 6 | Superwoman
7 | 9 | 4 7 | Lex Luther
8 | 15 | 9 8 | GodMaster
9 | 17 | 15 9 | Demon
10 | 16 | 15 10 | balou
11 | 6 | 5 11 | Superman
12 | 8 | 5 12 | xman
13 | 7 | 6 13 | hulk
14 | 13 | 8 14 | ironman
15 | 14 | 8 15 | aquaman
16 | 19 | 13 16 | wonderwoman
17 | 20 | 19 17 | batman
18 | 21 | 19 18 | robin
19 | 18 | 14 19 | tiger
20 | 1 | NULL 20 | oscar
这是我到目前为止的部分：
SELECT dbo.UserFamily.id, dbo.UserFamily.user_id, dbo.[User].name AS capo,
dbo.UserFamily.parent_id AS Capo_id, NULL AS underboss,
NULL AS underboss_id, NULL AS consigliere, NULL AS consigliere_id,
NULL AS godfather, NULL AS godfather_id
FROM dbo.UserFamily
INNER JOIN dbo.[User] ON dbo.UserFamily.parent_id = dbo.[User].id_user
ORDER BY dbo.UserFamily.user_id
这是我试图制作的视图（不知何故，user_id 1丢失，我认为与没有parent_id的事实有关）：
id | user_id | capo | capo_id | underboss | underboss_id | etc. | etc.
-----------------------------------------------------------------------------
1 | 2 | Victor | 1 | NULL | NULL | NULL
2 | 3 | Victor | 1 | NULL | NULL | NULL
3 | 4 | Victor | 1 | NULL | NULL | NULL
11 | 6 | Martin | 5 | Lucifer | 2 | Ewald
13 | 7 | Superwoman | 6 | Martin | 5 | Lucifer | 2
12 | 8 | Martin | 5 | Martin | 5 | Lucifer | 2
但我怎么得到其他的呢？像underboss等一直到教父？这是家庭结构：
soldier (the user)
capo
underboss
consigliere
godfather
答案 0 :(得分：2)
递归CTE是构建这样的层次结构的技巧。这是将层次结构渲染为字符串的一个很酷的技巧：
;WITH cteBosses AS
(
SELECT id_user, CAST(null AS VARCHAR(100)) AS parent_name, uf.parent_id,
name, 1 AS level, cast(name AS VARCHAR(MAX)) AS hierarchy
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
WHERE uf.parent_id IS NULL
UNION ALL
SELECT u.id_user, CAST(c.name AS VARCHAR(100)) AS parent_name, uf.parent_id,
u.name, c.level + 1 AS level, cast(c.hierarchy + ' -> ' + u.name AS VARCHAR(MAX)) AS hierarchy
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
INNER JOIN cteBosses c ON uf.parent_id = c.id_user
)
SELECT *
FROM cteBosses
ORDER BY hierarchy
输出：
id_user parent name parent_id level hierarchy
1 NULL Victor 1 1 Victor
3 Victor Hellboy 1 2 Victor -> Hellboy
2 Victor Lucifer 2 2 Victor -> Lucifer
5 Lucifer Martin 5 3 Victor -> Lucifer -> Martin
8 Martin GodMaster 8 4 Victor -> Lucifer -> Martin -> GodMaster
13 GodMaster hulk 13 5 Victor -> Lucifer -> Martin -> GodMaster -> hulk
19 hulk tiger 19 6 Victor -> Lucifer -> Martin -> GodMaster -> hulk -> tiger
etc.
阅读Steve Stedman的“Common Table Expressions Joes 2 Pros”，了解许多伟大的CTE示例。 http://stevestedman.com/2013/06/cte-scope/
要生成所需的输出，您可以根据需要多次将CTE连接到自身，如下所示：
;WITH cteBosses AS
(
SELECT id_user, CAST(null AS VARCHAR(100)) AS parent_name, uf.parent_id,
name, 1 AS level, cast(name AS VARCHAR(MAX)) AS hierarchy
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
WHERE uf.parent_id IS NULL
UNION ALL
SELECT u.id_user, CAST(c.name as VARCHAR(100)) AS parent_name, uf.parent_id,
u.name, c.level + 1 AS level, cast(c.hierarchy + ' -> ' + u.name AS VARCHAR(MAX)) AS hierarchy
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
INNER JOIN cteBosses c ON uf.parent_id = c.id_user
)
SELECT c1.id_user, c1.name, c2.id_user as capo_id, c2.name as capo,
c3.id_user as underboss_id, c3.name as underboss,
c4.id_user as next_boss_id, c4.name as next_boss
FROM cteBosses c1
LEFT JOIN cteBosses c2 ON c1.parent_id = c2.id_user
LEFT JOIN cteBosses c3 ON c2.parent_id = c3.id_user
LEFT JOIN cteBosses c4 ON c3.parent_id = c4.id_user
id_user name capo_id capo underboss_id underboss next_boss_id next_boss
1 Victor NULL NULL NULL NULL NULL NULL
2 Lucifer 1 Victor NULL NULL NULL NULL
3 Hellboy 1 Victor NULL NULL NULL NULL
4 Spiderman 1 Victor NULL NULL NULL NULL
9 Demon 4 Spiderman 1 Victor NULL NULL
10 balou 4 Spiderman 1 Victor NULL NULL
15 aquaman 9 Demon 4 Spiderman 1 Victor
16 wonderwoman 15 aquaman 9 Demon 4 Spiderman
17 batman 15 aquaman 9 Demon 4 Spiderman
要回答有关添加CurrentPosition列的后续问题，您可以嵌套另一个CTE并将其用于EXISTS子句：
;WITH cteBosses AS
(
SELECT id_user, CAST(null AS VARCHAR(100)) AS parent_name, uf.parent_id,
name, 1 AS level, cast(name AS VARCHAR(MAX)) AS hierarchy, id_user as top_parent, name as top_parent_name
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
WHERE uf.parent_id IS NULL
UNION ALL
SELECT u.id_user, CAST(c.name as VARCHAR(100)) AS parent_name, uf.parent_id,
u.name, c.level + 1 AS level, cast(c.hierarchy + ' -> ' + u.name AS VARCHAR(MAX)) AS hierarchy,
c.top_parent, c.top_parent_name
FROM [dbo.user] u
INNER JOIN dbo.userFamily uf ON u.id_user = uf.user_id
INNER JOIN cteBosses c ON uf.parent_id = c.id_user
)
, cteHierarchy AS
(
SELECT c1.id_user, c1.name, c2.id_user as capo_id, c2.name as capo,
c3.id_user as underboss_id, c3.name as underboss,
c4.id_user as next_boss_id, c4.name as next_boss
FROM cteBosses c1
LEFT JOIN cteBosses c2 ON c1.parent_id = c2.id_user
LEFT JOIN cteBosses c3 ON c2.parent_id = c3.id_user
LEFT JOIN cteBosses c4 ON c3.parent_id = c4.id_user
)
select id_user, name, --*,
case
when exists (select * from cteHierarchy cInner where next_boss = cOuter.name) then 'next_boss'
when exists (select * from cteHierarchy cInner where underboss = cOuter.name) then 'underboss'
when exists (select * from cteHierarchy cInner where capo = cOuter.name) then 'capo'
else 'regular_dude'
end as CurrentPosition
from cteHierarchy cOuter
id_user name CurrentPosition
1 Victor next_boss
4 Spiderman next_boss
9 Demon underboss
您可以继续添加这样的CTE级别 - CTE可以引用之前的CTE（但不是之后）。因此，您基本上为查询输出提供表别名，然后使用EXISTS计算CurrentPosition值。