如何从层次结构的任何级别使用 SQL 服务器获取层次结构查询中的根
How to get the root in a hierarchy query using SQL Server from any level of Hierarchy
我想从任何级别的数据中获取层次结构的最顶层祖先(根)。
以下是我的table.
CREATE TABLE #SMGROUP (ID INT NOT NULL, GRP NVARCHAR(40), GRPCLASS INT, PARENTGRP NVARCHAR(40), PARENTGRPCLASS INT)
INSERT INTO #SMGROUP VALUES (1, 'A', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'B', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'C', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'A.1', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'A.2', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'A.3', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'B.1', 2, 'B',1)
INSERT INTO #SMGROUP VALUES (1, 'B.2', 2, 'B',1)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3', 3, 'A.3',2)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3.3', 4, 'A.3.3',3)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3.3.1', 5, 'A.3.3.3',4)
INSERT INTO #SMGROUP VALUES (1, 'B.1.2', 3, 'B.1',2)
INSERT INTO #SMGROUP VALUES (1, 'B.2.1', 3, 'B.2', 2)
SELECT * FROM #SMGROUP
如果我提供 'A.1' 作为输入,我希望得到 - 'A' 的值,如果我提供 [=],return 的值也将是 'A' 22=] 作为输入。如果参数是 'A.3.3.3.1'
,return 也将是 'A'
我已经写了一些这样的东西,但我不知道如何继续。
;WITH items AS (
SELECT G.GRP ,CAST('' AS NVARCHAR(30)) AS ParentGroup,
0 AS Level
FROM #SMGROUP G
WHERE G.PARENTGRP IS NULL
UNION ALL
SELECT G.GRP, CAST(G.PARENTGRP AS NVARCHAR(30)) AS ParentGroup
, Level + 1
FROM #SMGROUP G
INNER JOIN items itms ON itms.GRP = G.PARENTGRP
)
SELECT * FROM items
你的方向是正确的,你只需要最后一击。
您 "reverse" 不是使用从根节点遍历到叶节点的 "standard" 递归 cte,而是从输入节点遍历回到根节点的过程。
然后它只是 top 1
和 order by
子句中的 level desc
:
DECLARE @GRP NVARCHAR(40) = 'A.3.3.3.1';
WITH items AS (
SELECT G.GRP,
ISNULL(G.PARENTGRP, '') AS ParentGroup,
0 AS Level
FROM #SMGROUP G
WHERE G.GRP = @GRP
UNION ALL
SELECT G.GRP,
G.PARENTGRP,
Level + 1
FROM #SMGROUP G
INNER JOIN items itms
ON itms.ParentGroup = G.GRP
)
SELECT TOP 1 Grp
FROM items
ORDER BY Level DESC
我想从任何级别的数据中获取层次结构的最顶层祖先(根)。 以下是我的table.
CREATE TABLE #SMGROUP (ID INT NOT NULL, GRP NVARCHAR(40), GRPCLASS INT, PARENTGRP NVARCHAR(40), PARENTGRPCLASS INT)
INSERT INTO #SMGROUP VALUES (1, 'A', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'B', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'C', 1, NULL,NULL)
INSERT INTO #SMGROUP VALUES (1, 'A.1', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'A.2', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'A.3', 2, 'A',1)
INSERT INTO #SMGROUP VALUES (1, 'B.1', 2, 'B',1)
INSERT INTO #SMGROUP VALUES (1, 'B.2', 2, 'B',1)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3', 3, 'A.3',2)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3.3', 4, 'A.3.3',3)
INSERT INTO #SMGROUP VALUES (1, 'A.3.3.3.1', 5, 'A.3.3.3',4)
INSERT INTO #SMGROUP VALUES (1, 'B.1.2', 3, 'B.1',2)
INSERT INTO #SMGROUP VALUES (1, 'B.2.1', 3, 'B.2', 2)
SELECT * FROM #SMGROUP
如果我提供 'A.1' 作为输入,我希望得到 - 'A' 的值,如果我提供 [=],return 的值也将是 'A' 22=] 作为输入。如果参数是 'A.3.3.3.1'
,return 也将是 'A'我已经写了一些这样的东西,但我不知道如何继续。
;WITH items AS (
SELECT G.GRP ,CAST('' AS NVARCHAR(30)) AS ParentGroup,
0 AS Level
FROM #SMGROUP G
WHERE G.PARENTGRP IS NULL
UNION ALL
SELECT G.GRP, CAST(G.PARENTGRP AS NVARCHAR(30)) AS ParentGroup
, Level + 1
FROM #SMGROUP G
INNER JOIN items itms ON itms.GRP = G.PARENTGRP
)
SELECT * FROM items
你的方向是正确的,你只需要最后一击。
您 "reverse" 不是使用从根节点遍历到叶节点的 "standard" 递归 cte,而是从输入节点遍历回到根节点的过程。
然后它只是 top 1
和 order by
子句中的 level desc
:
DECLARE @GRP NVARCHAR(40) = 'A.3.3.3.1';
WITH items AS (
SELECT G.GRP,
ISNULL(G.PARENTGRP, '') AS ParentGroup,
0 AS Level
FROM #SMGROUP G
WHERE G.GRP = @GRP
UNION ALL
SELECT G.GRP,
G.PARENTGRP,
Level + 1
FROM #SMGROUP G
INNER JOIN items itms
ON itms.ParentGroup = G.GRP
)
SELECT TOP 1 Grp
FROM items
ORDER BY Level DESC