Question

假设我有一个递归表（例如，有经理的员工）和一个大小为0..n的ID列表。如何找到这些ID的最低公共父级？

例如，如果我的表格如下：

Id | ParentId
---|---------
 1 |     NULL
 2 |        1
 3 |        1
 4 |        2
 5 |        2
 6 |        3
 7 |        3
 8 |        7

然后以下几组ID导致以下结果（第一个是角落情况）：

[]      => 1 (or NULL, doesn't really matter)
[1]     => 1
[2]     => 2
[1,8]   => 1
[4,5]   => 2
[4,6]   => 1
[6,7,8] => 3

怎么做？

编辑：请注意，在所有情况下，父级都不是正确的术语。它是树中所有路径中的最低公共节点。最低公共节点也可以是节点本身（例如，在[1,8] => 1的情况下，节点1不是节点1的父节点，而是节点1本身的节点。 / p>

亲切的问候，罗纳德

Answer 1

这是一种做法;它使用递归CTE来查找节点的祖先，并在输入值上使用“CROSS APPLY”来获得共同的祖先;您只需更改@ids（表变量）中的值：

----------------------------------------- SETUP
CREATE TABLE MyData (
   Id int NOT NULL,
   ParentId int NULL)

INSERT MyData VALUES (1,NULL)
INSERT MyData VALUES (2,1)
INSERT MyData VALUES (3,1)
INSERT MyData VALUES (4,2)
INSERT MyData VALUES (5,2)
INSERT MyData VALUES (6,3)
INSERT MyData VALUES (7,3)
INSERT MyData VALUES (8,7)

GO
CREATE FUNCTION AncestorsUdf (@Id int)
RETURNS TABLE
AS
RETURN (
    WITH Ancestors (Id, ParentId)
    AS (
        SELECT Id, ParentId
        FROM MyData
        WHERE Id = @Id
        UNION ALL
        SELECT md.Id, md.ParentId
        FROM MyData md
        INNER JOIN Ancestors a
          ON md.Id = a.ParentId
    )
    SELECT Id FROM Ancestors
);
GO
----------------------------------------- ACTUAL QUERY
DECLARE @ids TABLE (Id int NOT NULL)
DECLARE @Count int
-- your data (perhaps via a "split" udf)
INSERT @ids VALUES (6)
INSERT @ids VALUES (7)
INSERT @ids VALUES (8)

SELECT @Count = COUNT(1) FROM @ids
;
SELECT TOP 1 a.Id
FROM @ids
CROSS APPLY AncestorsUdf(Id) AS a
GROUP BY a.Id
HAVING COUNT(1) = @Count
ORDER BY a.ID DESC

如果节点没有严格提升，请更新：

CREATE FUNCTION AncestorsUdf (@Id int)
RETURNS @result TABLE (Id int, [Level] int)
AS
BEGIN
    WITH Ancestors (Id, ParentId, RelLevel)
    AS (
        SELECT Id, ParentId, 0
        FROM MyData
        WHERE Id = @Id
        UNION ALL
        SELECT md.Id, md.ParentId, a.RelLevel - 1
        FROM MyData md
        INNER JOIN Ancestors a
          ON md.Id = a.ParentId
    )

    INSERT @result
    SELECT Id, RelLevel FROM Ancestors

    DECLARE @Min int
    SELECT @Min = MIN([Level]) FROM @result

    UPDATE @result SET [Level] = [Level] - @Min

    RETURN
END
GO

和

SELECT TOP 1 a.Id
FROM @ids
CROSS APPLY AncestorsUdf(Id) AS a
GROUP BY a.Id, a.[Level]
HAVING COUNT(1) = @Count
ORDER BY a.[Level] DESC

Answer 2

在从Marc的回答中做出一些正确的思考和一些提示之后（谢谢），我自己提出了另一个解决方案：

DECLARE @parentChild TABLE (Id INT NOT NULL, ParentId INT NULL);
INSERT INTO @parentChild VALUES (1, NULL);
INSERT INTO @parentChild VALUES (2, 1);
INSERT INTO @parentChild VALUES (3, 1);
INSERT INTO @parentChild VALUES (4, 2);
INSERT INTO @parentChild VALUES (5, 2);
INSERT INTO @parentChild VALUES (6, 3);
INSERT INTO @parentChild VALUES (7, 3);
INSERT INTO @parentChild VALUES (8, 7);

DECLARE @ids TABLE (Id INT NOT NULL);
INSERT INTO @ids VALUES (6);
INSERT INTO @ids VALUES (7);
INSERT INTO @ids VALUES (8);

DECLARE @count INT;
SELECT @count = COUNT(1) FROM @ids;

WITH Nodes(Id, ParentId, Depth) AS
(
    -- Start from every node in the @ids collection.
    SELECT pc.Id , pc.ParentId , 0 AS DEPTH
    FROM @parentChild pc
    JOIN @ids i ON pc.Id = i.Id

    UNION ALL

    -- Recursively find parent nodes for each starting node.
    SELECT pc.Id , pc.ParentId , n.Depth - 1
    FROM @parentChild pc
    JOIN Nodes n ON pc.Id = n.ParentId
)
SELECT n.Id
FROM Nodes n
GROUP BY n.Id
HAVING COUNT(n.Id) = @count
ORDER BY MIN(n.Depth) DESC

它现在返回从最低公共父节点到根节点的整个路径，但这需要在select中添加TOP 1。

在递归SQL表中查找最低公共父级

2 个答案: