在SQL Server中执行递归自联接的最简单方法是什么?我有一张这样的桌子:
PersonID | Initials | ParentID
1 CJ NULL
2 EB 1
3 MB 1
4 SW 2
5 YT NULL
6 IS 5
我希望能够获得仅与特定人员开始的层次结构相关的记录。所以如果我通过PersonID = 1请求CJ的层次结构,我会得到:
PersonID | Initials | ParentID
1 CJ NULL
2 EB 1
3 MB 1
4 SW 2
对于EB,我会得到:
PersonID | Initials | ParentID
2 EB 1
4 SW 2
我有点卡在这上面,除了基于一堆连接的固定深度响应之外,我无法想到如何做到这一点。这会发生,因为我们不会有很多级别,但我想做得恰到好处。
谢谢!克里斯。
答案 0 :(得分:96)
WITH q AS
(
SELECT *
FROM mytable
WHERE ParentID IS NULL -- this condition defines the ultimate ancestors in your chain, change it as appropriate
UNION ALL
SELECT m.*
FROM mytable m
JOIN q
ON m.parentID = q.PersonID
)
SELECT *
FROM q
通过添加排序条件,您可以保留树顺序:
WITH q AS
(
SELECT m.*, CAST(ROW_NUMBER() OVER (ORDER BY m.PersonId) AS VARCHAR(MAX)) COLLATE Latin1_General_BIN AS bc
FROM mytable m
WHERE ParentID IS NULL
UNION ALL
SELECT m.*, q.bc + '.' + CAST(ROW_NUMBER() OVER (PARTITION BY m.ParentID ORDER BY m.PersonID) AS VARCHAR(MAX)) COLLATE Latin1_General_BIN
FROM mytable m
JOIN q
ON m.parentID = q.PersonID
)
SELECT *
FROM q
ORDER BY
bc
通过更改ORDER BY
条件,您可以更改兄弟姐妹的顺序。
答案 1 :(得分:22)
使用CTE你可以这样做
DECLARE @Table TABLE(
PersonID INT,
Initials VARCHAR(20),
ParentID INT
)
INSERT INTO @Table SELECT 1,'CJ',NULL
INSERT INTO @Table SELECT 2,'EB',1
INSERT INTO @Table SELECT 3,'MB',1
INSERT INTO @Table SELECT 4,'SW',2
INSERT INTO @Table SELECT 5,'YT',NULL
INSERT INTO @Table SELECT 6,'IS',5
DECLARE @PersonID INT
SELECT @PersonID = 1
;WITH Selects AS (
SELECT *
FROM @Table
WHERE PersonID = @PersonID
UNION ALL
SELECT t.*
FROM @Table t INNER JOIN
Selects s ON t.ParentID = s.PersonID
)
SELECT *
FROm Selects
答案 2 :(得分:4)
Quassnoi查询,包含对大表的更改。有更多孩子的父母然后10:格式化为str(5)row_number()
WITH q AS ( SELECT m.*, CAST(str(ROW_NUMBER() OVER (ORDER BY m.ordernum),5) AS VARCHAR(MAX)) COLLATE Latin1_General_BIN AS bc FROM #t m WHERE ParentID =0 UNION ALL SELECT m.*, q.bc + '.' + str(ROW_NUMBER() OVER (PARTITION BY m.ParentID ORDER BY m.ordernum),5) COLLATE Latin1_General_BIN FROM #t m JOIN q ON m.parentID = q.DBID ) SELECT * FROM q ORDER BY bc
答案 3 :(得分:2)
SQL 2005或更高版本,根据显示的示例,CTE是标准的方法。
SQL 2000,你可以使用UDF来实现 -
CREATE FUNCTION udfPersonAndChildren
(
@PersonID int
)
RETURNS @t TABLE (personid int, initials nchar(10), parentid int null)
AS
begin
insert into @t
select * from people p
where personID=@PersonID
while @@rowcount > 0
begin
insert into @t
select p.*
from people p
inner join @t o on p.parentid=o.personid
left join @t o2 on p.personid=o2.personid
where o2.personid is null
end
return
end
(这将在2005年运作,它不是标准的做法。那就是说,如果你发现更容易的工作方式,可以用它来运行)
如果您确实需要在SQL7中执行此操作,您可以在sproc中大致执行上述操作但无法从中进行选择 - SQL7不支持UDF。
答案 4 :(得分:0)
检查以下内容有助于理解CTE递归的概念
DECLARE
@startDate DATETIME,
@endDate DATETIME
SET @startDate = '11/10/2011'
SET @endDate = '03/25/2012'
; WITH CTE AS (
SELECT
YEAR(@startDate) AS 'yr',
MONTH(@startDate) AS 'mm',
DATENAME(mm, @startDate) AS 'mon',
DATEPART(d,@startDate) AS 'dd',
@startDate 'new_date'
UNION ALL
SELECT
YEAR(new_date) AS 'yr',
MONTH(new_date) AS 'mm',
DATENAME(mm, new_date) AS 'mon',
DATEPART(d,@startDate) AS 'dd',
DATEADD(d,1,new_date) 'new_date'
FROM CTE
WHERE new_date < @endDate
)
SELECT yr AS 'Year', mon AS 'Month', count(dd) AS 'Days'
FROM CTE
GROUP BY mon, yr, mm
ORDER BY yr, mm
OPTION (MAXRECURSION 1000)