做一个递归自连接最简单的方法？

Question

什么是做在SQL Server中的递归自连接的最简单的方法？我有一个表是这样的：

PersonID | Initials | ParentID
1          CJ         NULL
2          EB         1
3          MB         1
4          SW         2
5          YT         NULL
6          IS         5

和我希望能够得到不仅关系到开始与特定的人的层次来记录。所以，如果我要求CJ的层次由是PersonID = 1，我会得到：

PersonID | Initials | ParentID
1          CJ         NULL
2          EB         1
3          MB         1
4          SW         2

而对于EB的我会得到：

PersonID | Initials | ParentID
2          EB         1
4          SW         2

我有点憋屈这可能怎么也想不到基于一堆加入从固定深入响应分开做。因为它发生，因为我们不会有很多的水平，但我希望做正确呢？这会怎么做。

谢谢！克里斯。

Answer 1

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，您可以保存树顺序：

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

通过改变ORDER BY条件可以改变兄弟姐妹的排序。

Answer 2

使用CTE你能做到这样

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

Answer 3

在Quassnoi查询与大表的改变。具有更多的孩子的父母然后10：格式化为STR（5）的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

Answer 4

SQL 2005或更高，的CTE是去按照所示的实施例中的标准方法。

SQL 2000，您使用UDF的可以做到这一点 -

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

（将在2005年的工作，它只是没有做它的标准方式。这就是说，如果你发现了更简单的方式工作，它运行）

如果你真的需要做到这一点SQL7中，你可以做大致一个存储过程上面，但不能从它选择 - SQL7不支持的UDF

Answer 5

检查以下，以帮助理解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)