Question

我有一个父子关系表，如下所示。我想像所有祖先和父母一样检索父或子ID的所有记录，如果可能的话，还要检查深度。例如，我想找到D的族，它将返回前14行，因为它们都属于同一族。可能有几套这样的家庭。我想与一个成员查询，并希望获得全家记录。是否可以使用 CTE 实现此目的？根据表记录的家庭结构：

                      A
                     / \
                    B   C   G   J
                   /     \ / \ / \
              M   D       E   H   K
             / \ /             \ / \
            N   F               I   L


                 R
                 |
                 S   U
                  \ /
                   T

请帮忙。表格如下：

   Parent   Child
    ------  ------
    A            B
    A            C
    B            D
    D            F
    M            F
    M            N
    C            E
    G            E
    G            H
    J            H
    J            K
    H            I
    K            I
    K            L
    R            S
    S            T
    U            T

谢谢，

Himadri

Answer 1

我将使用递归的字符串限制器发布我的解决方案。我想知道这样做对于一个大图表这样做是否有好处，但我认为它毕竟不是那么糟糕。基本上我是递归地处理关系，记录让我走到每一步的“路径”，并检查路径，以便当我达到循环关系时我可以停止。

对于任何想要尝试的人，我都会发布表格的创建和填写：

create table nodes
(
    p char, 
    c char
)
insert into nodes (p, c)
values
    ('A', 'B'), ('A', 'C'), ('B', 'D'), ('D', 'F'), ('M', 'F'), ('M', 'N'), ('C', 'E'),
    ('G', 'E'), ('G', 'H'), ('J', 'H'), ('J', 'K'), ('H', 'I'), ('K', 'I'), ('K', 'L'),
    ('R', 'S'), ('S', 'T'), ('U', 'T')

以下是查询：

declare @let char = 'D';
with cte as
(
    --get the node itself if it exists in the table at all
    select 
        top 1 @let as letter, '' as rec_path 
    from 
        nodes 
    where 
        c = @let or p = @let 

    union all
    -- get the direct relations of the node as a starting point
    select 
        case when c = @let then p else c end as letter, 
        cast(@let as varchar(max)) as rec_path
    from 
        nodes 
    where 
        c = @let or p = @let

    union all
    -- get all of the relations recursively until you reach a node you already processed
    select 
        case when c = cte.letter then p else c end as letter, 
        rec_path + cte.letter as rec_path
    from 
        cte
        join nodes on 
            (cte.letter = nodes.c and charindex(cast(nodes.p as varchar(1)), rec_path, 1) = 0 
            or (cte.letter = nodes.p and charindex(cast(nodes.c as varchar(1)), rec_path, 1) = 0))
)
select 
    distinct letter
from 
    cte

我希望它会有用。我意识到你的数据实际上并不是由字母组成，但是同样可以使用id-s再次使用字符串路径，甚至是xml。

Answer 2

此解决方案仅适用于非循环图，因为递归会在循环时中断，将数据集视为提供...

create table nodes
(
    p char, 
    c char
)
insert into nodes (p, c)
values
    ('A', 'B'), ('A', 'C'), ('B', 'D'), ('D', 'F'), ('M', 'F'), ('M', 'N'), ('C', 'E'),
    ('G', 'E'), ('G', 'H'), ('J', 'H'), ('J', 'K'), ('H', 'I'), ('K', 'I'), ('K', 'L'),
    ('R', 'S'), ('S', 'T'), ('U', 'T')

GO

我们可以获得前向树父 - ＆gt;通过递归原始节点记录来显示子项

CREATE VIEW dbo.Tree
AS
    WITH Hierarchy(r, p, c, [Level])
    AS 
    (
        SELECT p AS r,
               p,
               c,
               0 AS [Level]
        FROM dbo.nodes
        UNION ALL
        SELECT n.p AS r,
               t.p,
               t.c,
               t.[Level] + 1
        FROM Hierarchy t
        INNER JOIN dbo.nodes n ON n.c = t.r
        AND n.p != t.p
    )
    SELECT r, p, c, [Level]
    FROM Hierarchy

然后给出结果

r   p   c   Level
A   A   B   0
A   A   C   0
A   C   E   1
A   D   F   2
A   B   D   1
B   D   F   1
B   B   D   0
C   C   E   0
D   D   F   0
G   G   E   0
G   G   H   0
G   H   I   1
H   H   I   0
J   H   I   1
J   K   L   1
J   K   I   1
J   J   H   0
J   J   K   0
K   K   I   0
K   K   L   0
M   M   F   0
M   M   N   0
R   R   S   0
R   S   T   1
S   S   T   0
U   U   T   0

我们可以通过另一种方式做到这一点，这样我们就可以从一个孩子一直走到它的父母那里

CREATE VIEW dbo.ReverseTree
AS
    WITH Hierarchy(r, c, p, [Level])
    AS 
    (
        SELECT c AS r,
               c,
               p,
               0 AS [Level]
        FROM dbo.nodes
        UNION ALL
        SELECT n.c AS r,
               t.c,
               t.p,
               t.[Level] + 1
        FROM Hierarchy t
        INNER JOIN dbo.nodes n ON n.p = t.r
        AND n.c != t.c
    )
    SELECT r, c, p, [Level]
    FROM Hierarchy

结果看起来像这样

r   c   p   Level
B   B   A   0
C   C   A   0
D   D   B   0
D   B   A   1
E   C   A   1
E   E   C   0
E   E   G   0
F   F   D   0
F   F   M   0
F   D   B   1
F   B   A   2
H   H   G   0
H   H   J   0
I   I   H   0
I   I   K   0
I   K   J   1
I   H   J   1
I   H   G   1
K   K   J   0
L   K   J   1
L   L   K   0
N   N   M   0
S   S   R   0
T   T   S   0
T   T   U   0
T   S   R   1

这对于像痕迹痕迹

这样的东西很方便

Answer 3

我找到了解决方案。但我在while循环中使用了 CTE 。如果有人有任何其他解决方案，请建议。正如我上面提到的包含家庭记录的表格，或者您可以说图表。我们将其命名为tbl_ParentChild。

这是我的代码：

Declare @Child varchar(10), @RowsEffected int
Set @Child='D'-----It is the member whose family we want to find

 CREATE Table #PrntChld (Parent varchar(10),Child varchar(10))
 Insert Into #PrntChld
 Select Parent,Child  from tbl_ParentChild MF
 Where MF.Child=@Child or MF.Parent=@Child

 Select @RowsEffected=Count(*) from #PrntChld

 While @RowsEffected>0 
 BEGIN
    ;WITH Prnt(Parent,Child)
        AS
        ( Select M.Parent,M.Child  from tbl_ParentChild M
            Inner Join #PrntChld F On F.Child=M.Child
          UNION ALL
          SELECT e.Parent,e.Child       
              FROM tbl_ParentChild AS E
              INNER JOIN  Prnt AS M                 
                  ON E.Child = M.Parent           
        ),
        PrntChld(Parent,Child)
        AS
        ( Select M.Parent,M.Child  from tbl_ParentChild M
            Inner Join (Select * from Prnt union Select * from #PrntChld) F On M.Parent=F.Parent
          UNION ALL
          SELECT e.Parent,e.Child       
              FROM tbl_ParentChild AS E
              INNER JOIN  PrntChld AS M                 
                  ON M.Child = E.Parent           
        )

    Insert Into #PrntChld
    Select distinct MF.* from PrntChld MF 
    Left Join #PrntChld T On T.Child =MF.Child and T.Parent  = MF.Parent    
    where T.Child is null
    Select @RowsEffected=@@ROWCOUNT

END

Select * from #PrntChld
drop table #PrntChld

与多父母的父子关系的CTE

3 个答案: