115

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

而且我希望能够获取仅与从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

除了基于一堆连接的固定深度响应之外，我对此有点坚持。这会发生，因为我们不会有很多级别，但我想正确地做到这一点。

谢谢！克里斯。

4

6 回答 6

127 
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。

于 2009-11-18T16:35:13.123 回答
26

使用 CTE，您可以这样做

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
于 2009-11-18T16:37:59.513 回答
5

对大表进行更改的 Quassnoi 查询。孩子多于 10 岁的父母：将 row_number() 格式化为 str(5)

与 q 作为
        (
        SELECT m.*, CAST(str(ROW_NUMBER() OVER (ORDER BY m.ordernum),5) AS VARCHAR(MAX)) COLLATE Latin1_General_BIN AS bc
        来自#tm
        父 ID =0
        联合所有
        选择 m.*, q.bc + '.' + str(ROW_NUMBER() OVER (PARTITION BY m.ParentID ORDER BY m.ordernum),5) 整理 Latin1_General_BIN
        来自#tm
        加入q
        开 m.parentID = q.DBID
        )
选择 *
来自 q
订购方式
        公元前
于 2009-11-18T17:42:15.047 回答
2

SQL 2005 或更高版本，CTE 是按照所示示例的标准方法。

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

如果您真的需要在 SQL7 中执行此操作，您可以在 sproc 中大致执行上述操作，但无法从中选择 - SQL7 不支持 UDF。

于 2009-11-18T17:18:06.440 回答
2

检查以下内容以帮助理解 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)
于 2018-09-28T10:00:02.103 回答
0 
DELIMITER $$

 

DROP PROCEDURE IF EXISTS `myprocDURENAME`$$

CREATE DEFINER=`root`@`%` PROCEDURE `myprocDURENAME`( IN grp_id VARCHAR(300))
BEGIN
   SELECT h.ID AS state_id,UPPER(CONCAT( `ACCNAME`,' [',b.`GRPNAME`,']')) AS state_name,h.ISACTIVE  FROM accgroup b JOIN (SELECT get_group_chield (grp_id) a) s ON FIND_IN_SET(b.ID,s.a) LEFT OUTER JOIN acc_head h ON b.ID=h.GRPID WHERE h.ID IS NOT NULL AND H.ISACTIVE=1;
    END$$

DELIMITER ;

////////////////////////

DELIMITER $$

 

DROP FUNCTION IF EXISTS `get_group_chield`$$

CREATE DEFINER=`root`@`%` FUNCTION `get_group_chield`(get_id VARCHAR(999)) RETURNS VARCHAR(9999) CHARSET utf8
BEGIN
     DECLARE idd VARCHAR(300);
   DECLARE get_val VARCHAR(300);
    DECLARE get_count INT;
SET idd=get_id;
   
SELECT  GROUP_CONCAT(id)AS t,COUNT(*) t1 INTO get_val,get_count FROM accgroup ag JOIN (SELECT idd AS n1) d ON FIND_IN_SET(ag.PRNTID,d.n1);
SELECT   COUNT(*) INTO get_count FROM accgroup WHERE PRNTID IN (idd);
 WHILE get_count >0 DO
 SET idd=CONCAT(idd,',', get_val); 
SELECT  GROUP_CONCAT(CONCAT('', id ,'' ))AS t,COUNT(*) t1 INTO get_val,get_count FROM accgroup ag JOIN (SELECT get_val AS n1) d ON FIND_IN_SET(ag.PRNTID,d.n1);
   END WHILE;
   RETURN idd;
-- SELECT id FROM acc_head WHERE GRPID  IN (idd);
    END$$

DELIMITER ;
于 2021-08-18T07:43:28.937 回答