具有对等关系的递归查询

Question

假设有一张关系表 (entity_id, 关系, related_id)

1, A, 2     
1, A, 3      
3, B, 5 
1, C, null 
12, C, 1 
100, C, null

我需要一个能提取所有相关行的查询。 比如我查询entity_id = 1，下面的行应该被拉出来

1, A, 2     
1, A, 3      
3, B, 5 
1, C, null 
12, C, 1 

其实如果我查询entity_id = 1, 2, 3, 5, or 12，结果集应该是一样的。

这与标准的经理-员工范式不同，因为没有层次结构。关系可以朝任何方向发展。

---

编辑 到目前为止发布的答案都没有奏效。

我想出了一个可行的解决方案。

我会将解决方案归功于能够将这个怪物清理成更优雅的东西的人。

with tab as ( 
-- union for reversals
 select id, entity_id, r.related_id, 1 level
 , cast('/' + cast(entity_id as varchar(1000)) + '/'  as varchar(1000)) path 
  from _entity_relation r 
  where not exists(select null from _entity_relation r2 where r2.related_id=r.entity_id) 
    or r.related_id is null 
 union
 select id, related_id, r.entity_id, 1 level
 , cast('/' + cast(related_id as varchar(1000)) + '/' as varchar(1000)) path 
  from _entity_relation r 
  where not exists(select null from _entity_relation r2 where r2.related_id=r.entity_id) 
    or r.related_id is null 

-- create recursive path
union all 
 select r.id, r.entity_id, r.related_id, tab.level+1
 , cast(tab.path + '/' + cast(r.entity_id as varchar(100)) + '/' + '/' + cast(r.related_id as varchar(1000)) + '/' as varchar(1000)) path 
  from _entity_relation r 
  join tab 
  on tab.related_id = r.entity_id   
) 

select x.id
    , x.entity_id
    ,pr.description as relation_description
    ,pt.first_name + coalesce(' ' + pt.middle_name,'') + ' ' + pt.last_name as relation_name
    ,CONVERT(CHAR(10), pt.birth_date, 101) as relation_birth_date   
from (

select entity_id, MAX(id) as id from (
select distinct tab.id, entity_id
from tab 
join( 
    select path 
    from tab  
    where entity_id=@in_entity_id
) p on p.path like tab.path + '%' or tab.path like p.path + '%'
union
select distinct tab.id, related_id
from tab 
join( 
    select path 
    from tab  
    where entity_id=@in_entity_id
) p on p.path like tab.path + '%' or tab.path like p.path + '%'
union
select distinct tab.id, entity_id
from tab 
join( 
    select path 
    from tab  
    where related_id=@in_entity_id
) p on p.path like tab.path + '%' or tab.path like p.path + '%'
union
select distinct tab.id, related_id
from tab 
join( 
    select path 
    from tab  
    where related_id=@in_entity_id
) p on p.path like tab.path + '%' or tab.path like p.path + '%'
) y
group by entity_id
) x
join _entity_relation pr on pr.id = x.id
join _entity pt on pt.id = x.entity_id
where x.entity_id <> @in_entity_id;

Answer 1

请注意您的数据，因为要完成您的任务，您必须避免循环引用。以下查询可以优化，但肯定会起作用

;with tab as (
 select entity_id, relationship, related_id, 1 level, cast('/' + cast(entity_id as varchar(1000)) as varchar(1000)) path
  from #r r
  where not exists(select null from #r r2 where r2.related_id=r.entity_id)
    or r.related_id is null
union all
 select r.entity_id, r.relationship, r.related_id, tab.level+1, cast(tab.path + '/' + cast(r.entity_id as varchar(100)) as varchar(1000)) path
  from #r r
  join tab
  on tab.related_id = r.entity_id  
)
select distinct tab.* 
    from tab
    join(
select path
    from tab 
    where entity_id=1) p
    on p.path like tab.path + '%' or tab.path like p.path + '%'

Answer 2

使用两个 CTE 的解决方案

我首先创建了一个具有双向关系的表，然后创建了一个递归 CTE，它使用这两种方式的结果来构建具有祖先路径的整个层次结构...

with both as
(
    select *, 0 as rev
    from t
    where related_id is not null

    union

    select *, 1
    from t
),
recurs as
(
    select *, cast('/' as varchar(100)) as anc
    from both
    where entity_id is null

    union all

    select b.*, cast(re.anc + cast(b.entity_id as varchar) + '/' as varchar(100))
    from both b
        join recurs re
        on (re.related_id = b.entity_id)
    where charindex('/'+cast(isnull(b.entity_id,'') as varchar)+'/', re.anc) = 0
)
select *
/*
    THIS ONE SHOULD BE USED TO RETURN TO ORIGINAL
    case when is_reverse = 1 then related_id else entity_id end as entity_id,
    relationship,
    case when is_reverse = 0 then related_id else entity_id end as related_id
*/
from recurs
where related_id = xXx or
      charindex('/'+cast(xXx as varchar)+'/', anc) != 0

用实际值替换xXx。

这个查询假定根元素是带有entity_id = null 的那个，所以它从那里构建整个递归。如果不是这种情况，您将不得不相应地进行更改。

我添加了循环检查，循环是 1,2,3,4,5,1 还是 1,2,3,4,5,3。 ..所以全部或部分循环。两者都可以。