[英]How to use recursive logic to return only the Root row in a sql table (SQL Server 2008 R2)
這是針對SQL Server 2008 R2的,我是SQL的新手,所以請盡可能具體。
Table1
內置了一些遞歸結構,其中ParentId
為Null
表示它是根,或者ParentId
是Table1
中另一行的Id
(表示其為子代)。
數據集示例:
Table1Id ParentId
--------------------------------------------
1 NULL
2 1
3 1
4 2
5 NULL
6 2
7 6
8 NULL
9 8
在上面的示例中,表格具有以下帶有3個根節點的樹結構:
Root 1 5 8
Child(teir1) 2 3 9
Child(teir2) 4 6
Child(tier3) 7
....
給定任何行ID,是否有辦法僅返回“根”行? 例如:
InputId ReturnedRowId
----------------------------
1 1
2 1
3 1
4 1
5 5
6 1
7 1
8 8
9 8
任何幫助,將不勝感激。
您可以使用CTE遍歷層次結構
IF OBJECT_ID('tempdb..#testData') IS NOT NULL
DROP TABLE #testData
CREATE TABLE #testData (
Table1Id INT
,ParentId INT NULL
)
INSERT INTO #testData ( Table1Id, ParentId )
VALUES
(1, NULL )
,(2, 1 )
,(3, 1 )
,(4, 2 )
,(5, NULL )
,(6, 2 )
,(7, 6 )
,(8, NULL )
,(9, 8 )
DECLARE @InputId INT
SET @InputId = 2 --<<--Change this as appropriate
;WITH cteTraverse
AS
(
SELECT
T.Table1Id, T.ParentId
FROM
#testData T
WHERE
Table1Id = @InputId
UNION ALL
SELECT
T1.Table1Id, T1.ParentId
FROM
#testData T1
INNER JOIN
cteTraverse T2 ON T1.Table1Id = T2.ParentId
)
SELECT
@InputId '@InputId', Table1Id 'ReturnedRowId'
FROM
cteTraverse
WHERE
ParentId IS NULL
這個查詢完成工作。
with CTE as
(
Select Table1ID as ID, Table1ID as Ancestor, 0 as level
from Table1
UNION ALL
Select ID, ParentID, level + 1
from Table1
inner join CTE on CTE.Ancestor = Table1.Table1ID
where ParentID is not NULL
)
,
R_only as
(
Select ID as ID, MAX(level) as max_level
from CTE
group by ID
)
select CTE.ID, Ancestor
from CTE inner join R_only on CTE.ID = R_only.ID and CTE.level = R_only.max_level
order by CTE.ID
這是一個腳本,可為您擁有的節點表找到根節點。 它能做什么:
level
的深度。 max_level
。 這是確定父級的深度。 CREATE TABLE #tree(table1_id INT PRIMARY KEY,parent_id INT);
INSERT INTO #tree(table1_id,parent_id)VALUES
(1,NULL),(2,1),(3,1),(4,2),(5,NULL),(6,2),(7,6),(8,NULL),(9,8);
;WITH cte_tr AS (
SELECT table1_id, parent_id, level=0
FROM #tree
UNION ALL
SELECT t_c.table1_id, t_p.parent_id, level=t_c.level+1
FROM cte_tr AS t_c
INNER JOIN #tree AS t_p ON
t_p.table1_id=t_c.parent_id
WHERE t_p.parent_id IS NOT NULL
),
cte_ml AS (
SELECT table1_id, max_level=MAX(level)
FROM cte_tr
GROUP BY table1_id
)
SELECT cte_tr.table1_id, root_node=ISNULL(cte_tr.parent_id,cte_tr.table1_id)
FROM cte_tr
INNER JOIN cte_ml ON
cte_ml.table1_id=cte_tr.table1_id AND
cte_ml.max_level=cte_tr.level
ORDER BY cte_tr.table1_id
DROP TABLE #tree;
結果:
+-----------+-----------+
| table1_id | root_node |
+-----------+-----------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 5 | 5 |
| 6 | 1 |
| 7 | 1 |
| 8 | 8 |
| 9 | 8 |
+-----------+-----------+
使用“開始於”和“事先連接”。 在此處查看此文檔
http://psoug.org/reference/connectby.html
我不確定除Oracle以外的其他數據庫是否可以使用“ connect by”。 看看這個。
您也可以嘗試with子句。 有人在這里嘗試過。
SQL Server中ORACLE的PRIOR CONNECT的仿真
與這樣的作品
with query1 as (select .... from .... where ....),
query2 as (select .... from .... where ....)
select ...
from query1 q1,
query2 q2
where q1.xxxxx = q2.xxxx
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