进行递归自联接的最简单方法？

在SQL Server中进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

而且我希望能够只获取与特定人员开始的层次结构相关的记录。因此，如果我通过PersonID = 1请求CJ的层次结构，则会得到：

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

除了基于一堆联接的固定深度响应之外，我对此深感困惑。这样做会发生，因为我们没有很多级别，但我想正确地做。

谢谢！克里斯。

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，可以保留树的顺序：

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。

使用CTE，您可以采用这种方式

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

Quassnoi查询更改了大表。多于10个孩子的父母：将row_number（）格式化为str（5）

带q AS 
        （
        SELECT m。*，CAST（str（ROW_NUMBER（）OVER（OR BY BY m.ordernum），5）AS VARCHAR（MAX））COLLICA Latin1_General_BIN AS bc
        来自#tm
        父母ID = 0
        全联盟
        选择m。*，q.bc +'。' + str（ROW_NUMBER（）OVER（PARTITION BY m.ParentID ORDER BY m.ordernum），5）COLLECTION Latin1_General_BIN
        来自#tm
        加入q
        开启m.parentID = q.DBID
        ）
选择 *
从q
订购
        公元前

按照所示示例，SQL 2005或更高版本是CTE的标准处理方式。

SQL 2000，您可以使用UDF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年生效，但这不是标准的操作方式。也就是说，如果您发现更简单的工作方式，请使用它）

如果确实需要在SQL7中执行此操作，则可以在sproc中大致执行上述操作，但不能从中选择-SQL7不支持UDF。

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)