[英]How to use recursive logic to return only the Root row in a sql table (SQL Server 2008 R2)
这是针对SQL Server 2008 R2的,我是SQL的新手,所以请尽可能具体。
Table1
内置了一些递归结构,其中ParentId
为Null
表示它是根,或者ParentId
是Table1
中另一行的Id
(表示其为子代)。
数据集示例:
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,表格具有以下带有3个根节点的树结构:
Root 1 5 8
Child(teir1) 2 3 9
Child(teir2) 4 6
Child(tier3) 7
....
给定任何行ID,是否有办法仅返回“根”行? 例如:
InputId ReturnedRowId
----------------------------
1 1
2 1
3 1
4 1
5 5
6 1
7 1
8 8
9 8
任何帮助,将不胜感激。
您可以使用CTE遍历层次结构
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
这个查询完成工作。
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
这是一个脚本,可为您拥有的节点表找到根节点。 它能做什么:
level
的深度。 max_level
。 这是确定父级的深度。 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;
结果:
+-----------+-----------+
| table1_id | root_node |
+-----------+-----------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 5 | 5 |
| 6 | 1 |
| 7 | 1 |
| 8 | 8 |
| 9 | 8 |
+-----------+-----------+
使用“开始于”和“事先连接”。 在此处查看此文档
http://psoug.org/reference/connectby.html
我不确定除Oracle以外的其他数据库是否可以使用“ connect by”。 看看这个。
您也可以尝试with子句。 有人在这里尝试过。
SQL Server中ORACLE的PRIOR CONNECT的仿真
与这样的作品
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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