我有一张带有parent和childid的表。
create table if not exists stack (
parent int,
child int
)
每个父母可以有多个孩子，每个孩子又可以有多个孩子。
insert into stack (parent, child) values
(1,2),
(2,3),
(3,4),
(4,5),
(5,6),
(6,7),
(7,8),
(8,9),
(9,null),
(1,7),
(7,8),
(8,9),
(9,null);
数据看起来像这样。
|parent|child|
|------|-----|
|1 |2 |
|2 |3 |
|3 |4 |
|4 |5 |
|5 |6 |
|6 |7 |
|7 |8 |
|8 |9 |
|9 |NULL |
|1 |7 |
|7 |8 |
|8 |9 |
|9 |NULL |
我想找到所有的孩子。我可以将递归 cte 与UNION ALL.
with recursive cte as (
select
child
from
stack
where
stack.parent = 1
union
select
stack.child
from
cte
left join stack on
cte.child = stack.parent
where
cte.child is not null
)
select * from cte;
这给了我想要达到的结果。
|child|
|-----|
|2 |
|7 |
|3 |
|8 |
|4 |
|9 |
|5 |
|NULL |
|6 |
但是我想包括深度/级别以及每个节点的路径。我可以使用不同的递归 cte 来做到这一点。
with recursive cte as (
select
parent,
child,
0 as level,
array[parent,
child] as path
from
stack
where
stack.parent = 1
union all
select
stack.parent,
stack.child,
cte.level + 1,
cte.path || stack.child
from
cte
left join stack on
cte.child = stack.parent
where
cte.child is not null
)
select * from cte;
这给了我这个数据。
|parent|child|level|path |
|------|-----|-----|--------------------|
|1 |2 |0 |{1,2} |
|1 |7 |0 |{1,7} |
|2 |3 |1 |{1,2,3} |
|7 |8 |1 |{1,7,8} |
|7 |8 |1 |{1,7,8} |
|3 |4 |2 |{1,2,3,4} |
|8 |9 |2 |{1,7,8,9} |
|8 |9 |2 |{1,7,8,9} |
|8 |9 |2 |{1,7,8,9} |
|8 |9 |2 |{1,7,8,9} |
|4 |5 |3 |{1,2,3,4,5} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|9 | |3 |{1,7,8,9,} |
|5 |6 |4 |{1,2,3,4,5,6} |
|6 |7 |5 |{1,2,3,4,5,6,7} |
|7 |8 |6 |{1,2,3,4,5,6,7,8} |
|7 |8 |6 |{1,2,3,4,5,6,7,8} |
|8 |9 |7 |{1,2,3,4,5,6,7,8,9} |
|8 |9 |7 |{1,2,3,4,5,6,7,8,9} |
|8 |9 |7 |{1,2,3,4,5,6,7,8,9} |
|8 |9 |7 |{1,2,3,4,5,6,7,8,9} |
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
|9 | |8 |{1,2,3,4,5,6,7,8,9,}|
我的问题是我有很多重复的数据。我想获得与UNION查询相同的结果，但具有级别和路径。
我试过类似的东西
where
cte.child is not null
and stack.parent not in (cte.parent)
或者
where
cte.child is not null
and not exists (select parent from cte where cte.parent = stack.parent)
但第一个不会改变任何东西，第二个返回错误。
ERROR: recursive reference to query "cte" must not appear within a subquery
有任何想法吗？非常感谢！
您的问题是表格数据不合适。例如，您的表包含 8 两次是 7 的直接子代的信息。我建议您删除重复数据并在对上实施唯一约束。
如果由于某种原因不能这样做，请在查询中使行不同：
with recursive
good_stack as (select distinct * from stack)
,cte as
(
select
parent,
child,
0 as level,
array[parent,
child] as path
from good_stack
where good_stack.parent = 1
union all
select
good_stack.parent,
good_stack.child,
cte.level + 1,
cte.path || good_stack.child
from cte
left join good_stack on cte.child = good_stack.parent
where cte.child is not null and good_stack.child is not null
)
select * from cte;
演示：https : //dbfiddle.uk/?rdbms=postgres_13&fiddle=acb1d7a1a1d26c3fd9caf0e7dedc12b2
（您也可以使列不可为空。条目 9|null 不添加任何信息。如果表缺少这些条目，则 9 仍然没有子项。）
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句