父子 SQL 递归 [英] Parent Child SQL Recursion

问题描述

我见过类似但不完全相同的请求.

I have seen similar but not Exactly the same requests.

如果我有下表

Parent  Child
1       2
1       3
4       3
5       1
6       1
5       7
8       9

我选择了1"我希望返回所有记录，其中一个是父级或子级，但也包括所有相关的父级和子级，例如行5、7"，因为 5 是1"的父级

I selected "1" I would expect back all records where one is the parent or child but also all related parents and children for instance row "5 , 7" because 5 is the parent of "1"

所以 1 的结果集将是

so the result set for 1 would be

Parent  Child
1       2
1       3
4       3
5       1
6       1
5       7

所以它会不包括行

So it would NOT include the row

Parent  Child
8       9

这是我目前所能达到的最接近的

This is as close as I can get so far

;WITH LinksDown AS (
   SELECT *
   FROM RecursiveTable
   WHERE Parent = 1
   UNION ALL
   SELECT rt.*
   FROM RecursiveTable rt
   JOIN LinksDown ld on ld.Child = rt.Parent
),

LinksUp AS (
   SELECT *
   FROM RecursiveTable
   WHERE Child = 1
   UNION ALL
   SELECT rt.*
   FROM RecursiveTable rt
   JOIN LinksUp lu on lu.Child = rt.Parent
) 

select distinct *
from LinksDown

Union All 
select distinct * from LinksUp

但这有以下输出，远非所需

But this has the following output which is far from whats needed

Parent  Child
1       2
1       3
1       2
1       3
5       1
6       1

推荐答案

这里有两种方法.第一种使用效率很低的 CTE.问题是在递归期间您无法检查结果集中的所有其他行.虽然您可以构建对给定行有贡献的行的列表，但您无法检查是否已经通过另一条路径到达该行.第二种方法使用循环一步一步地用关系填充表.这是一种比 CTE 更好的方法.

Here are two approaches. The first uses a CTE that is quite inefficient. The problem is that during recursion you cannot examine all of the other rows in the result set. While you can build a list of the rows that have contributed to a given row, you cannot check to see if you already reached that row via another path. The second approach uses a loop to populate a table with relations one step at a time. It is a much better method than the CTE.

留给读者练习:这两种方法是否会在树"中存在循环时终止，例如1 > 2 > 3 > 1?

Left as an exercise for the reader: Will the two methods terminate in the presence of a cycle in the "tree", e.g. 1 > 2 > 3 > 1?

-- Sample data.
declare @RecursiveTable as Table ( Parent Int, Child Int );
insert into @RecursiveTable ( Parent, Child ) values
  ( 1, 2 ), ( 1, 3 ),
  ( 4, 3 ),
  ( 5, 1 ),
  ( 6, 1 ),
  ( 5, 7 ),
  ( 8, 9 );
select * from @RecursiveTable;

-- Walk the tree with a recursive CTE.
--   NB: This is woefully inefficient since we cannot promptly detect
--   rows that have already been processed.
declare @Start as Int = 1;
with Pairs as (
  select Parent, Child, Cast( Parent as VarChar(10) ) + '/' + Cast( Child as VarChar(10) ) as Pair
    from @RecursiveTable ),
 Relations as (
  select Parent, Child, Cast( '|' + Pair + '|' as VarChar(1024) ) as Path
    from Pairs
    where Parent = @Start or Child = @Start
  union all
  select P.Parent, P.Child, Cast( R.Path + P.Pair + '|' as VarChar(1024) )
    from Relations as R inner join
      Pairs as P on P.Child = R.Parent or P.Parent = R.Child or
        P.Child = R.Child or P.Parent = R.Parent
    where CharIndex( '|' + P.Pair + '|', R.Path ) = 0
  )
  -- To see how terrible this is, try: select * from Relations
  select distinct Parent, Child
    from Relations
    order by Parent, Child;

-- Try again a loop to add relations to a working table.
declare @Relations as Table ( Parent Int, Child Int );
insert into @Relations
  select Parent, Child
    from @RecursiveTable
    where Parent = @Start or Child = @Start;
while @@RowCount > 0
  insert into @Relations
    select RT.Parent, RT.Child
      from @Relations as R inner join
        @RecursiveTable as RT on RT.Child = R.Child or RT.Parent = R.Parent or
          RT.Child = R.Parent or RT.Parent = R.Child
    except
    select Parent, Child
      from @Relations;
select Parent, Child
  from @Relations
  order by Parent, Child;

这篇关于父子 SQL 递归的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！