Sql 在复杂的层次结构上使用递归

我尝试将递归用于非常复杂的层次结构，如下所示：

Root    
|    
Second   
|   
Third   
|   
Leaf   

with recursive relations as 
(select parent_id, child_id, 1 as h_level, array[parent_id,child_id] as fullpath
 from public.entity_relations where parent_type = 4
union all 
 select c.parent_id, c.child_id, p.h_level+1, p.fullpath || c.child_id 
 from public.entity_relations c join relations p on p.child_id = c.parent_id)
select * from relations;

但如果没有第二个，也可能是这样：

Root       
|   
Third   
|   
Leaf

我的SQL查询如下所示：

Root    
|    
Second   
|   
Third   
|   
Leaf   

with recursive relations as 
(select parent_id, child_id, 1 as h_level, array[parent_id,child_id] as fullpath
 from public.entity_relations where parent_type = 4
union all 
 select c.parent_id, c.child_id, p.h_level+1, p.fullpath || c.child_id 
 from public.entity_relations c join relations p on p.child_id = c.parent_id)
select * from relations;

下面是一个带有关系表和递归查询的SQLFIDLE链接：

问题是递归返回这个层次结构的子路径。 我只对每个层次结构的最完整路径感兴趣，在本例中，仅对以下记录感兴趣：

BB  CC  2   A,BB,CC
C   D   3   A,B,C,D

注意：我知道子路径对于递归是必要的，

---

因此，我正在寻找一种在递归之后过滤掉冗余记录的方法

在最终查询中，您只能选择没有子节点的叶子节点：

with recursive relations as (
    select 
        parent_id, 
        child_id, 
        1 as h_level, 
        array[parent_id,child_id] as fullpath
    from entity_relations where parent_type = 4
union all 
    select 
        c.parent_id, 
        c.child_id, 
        p.h_level+1, 
        p.fullpath || c.child_id 
    from entity_relations c 
    join relations p on p.child_id = c.parent_id
)
select *
from relations r
where not exists (
    select from entity_relations e
    where e.parent_id = r.child_id
)

 parent_id | child_id | h_level | fullpath
-----------+----------+---------+-----------
 C         | D        |       2 | {A,C,D}
 BB        | CC       |       2 | {A,BB,CC}
 C         | D        |       3 | {A,B,C,D}
(3 rows)

事实证明，有两条路径通向叶D。您可以根据级别选择其中一条路径

...
select distinct on (child_id) *
from relations r
where not exists (
    select from entity_relations e
    where e.parent_id = r.child_id
)
order by child_id, h_level desc

 parent_id | child_id | h_level | fullpath
-----------+----------+---------+-----------
 BB        | CC       |       2 | {A,BB,CC}
 C         | D        |       3 | {A,B,C,D}
(2 rows)    

---

在最终查询中，只能选择没有子节点的叶子节点：

with recursive relations as (
    select 
        parent_id, 
        child_id, 
        1 as h_level, 
        array[parent_id,child_id] as fullpath
    from entity_relations where parent_type = 4
union all 
    select 
        c.parent_id, 
        c.child_id, 
        p.h_level+1, 
        p.fullpath || c.child_id 
    from entity_relations c 
    join relations p on p.child_id = c.parent_id
)
select *
from relations r
where not exists (
    select from entity_relations e
    where e.parent_id = r.child_id
)

 parent_id | child_id | h_level | fullpath
-----------+----------+---------+-----------
 C         | D        |       2 | {A,C,D}
 BB        | CC       |       2 | {A,BB,CC}
 C         | D        |       3 | {A,B,C,D}
(3 rows)

事实证明，有两条路径通向叶D。您可以根据级别选择其中一条路径

...
select distinct on (child_id) *
from relations r
where not exists (
    select from entity_relations e
    where e.parent_id = r.child_id
)
order by child_id, h_level desc

 parent_id | child_id | h_level | fullpath
-----------+----------+---------+-----------
 BB        | CC       |       2 | {A,BB,CC}
 C         | D        |       3 | {A,B,C,D}
(2 rows)    

---

如果我理解正确，您希望：

with recursive relations as (
      select parent_id, child_id, 1 as h_level, array[parent_id,child_id] as fullpath
      from public.entity_relations
      where parent_type = 4
      union all 
      select c.parent_id, c.child_id, p.h_level+1, p.fullpath || c.child_id 
      from public.entity_relations c join
           relations p
           on p.child_id = c.parent_id
     )
select r.*
from relations r
where not exists (select 1
                  from relations r2
                  where r2.fullpath @> r.fullpath and
                        r2.fullpath <> r.fullpath
                 );

---

如果我理解正确，您希望：

with recursive relations as (
      select parent_id, child_id, 1 as h_level, array[parent_id,child_id] as fullpath
      from public.entity_relations
      where parent_type = 4
      union all 
      select c.parent_id, c.child_id, p.h_level+1, p.fullpath || c.child_id 
      from public.entity_relations c join
           relations p
           on p.child_id = c.parent_id
     )
select r.*
from relations r
where not exists (select 1
                  from relations r2
                  where r2.fullpath @> r.fullpath and
                        r2.fullpath <> r.fullpath
                 );

---

您希望返回什么结果集？您希望返回什么结果集？