Question

我有两张桌子：

locations：

locationid   parentid
  1
  2              1
  3              1
  4              2
  5              4
  8              1

approvelog：

locationid  approved
   4          True
   8          True

我需要编写一个查询，该查询提供特定位置的所有子位置，但如果locationid被批准，则忽略它及其子项，即使这些子项未获批准也是如此。

简单来说，当遇到具有approved=True的locationid时，忽略它及其所有子项（意味着停止递归此分支）。

例如：

我希望获得locationid=2：
```
2
```
我希望获得locationid=8：
```
Nothing
```
我希望获得locationid=1：
```
1,2,3
```
4被批准所以忽略它和它的孩子。 8被批准，所以忽略它。

这是我的代码：

with recursive location_tree as (
   select  locationid, parentid
   from locations
   where locationid = 1 and not approved
   union all
   select child.locationid, child.parentid
   from locations child
   join location_tree parent on parent.locationid = child.parentid
   where child.active
)
select array_agg(locationid) 
from location_tree

这只是给出一个位置列表。

基本上我需要的是递归的“停止条件”。如何更改它？

有人可以帮忙吗？

Answer 1

这是我在PostgreSQL中用来让父母及其未经批准的孩子认为父母本身也未被批准的内容：

WITH recursive location_tree AS (
  --PARENT
  SELECT
    p.locationid,
    p.parentid
  FROM
    locations p
  LEFT JOIN
    approvelog pa ON pa.locationid = p.locationid AND pa.approved = TRUE
  WHERE
    p.locationid = 1
    AND pa.locationid IS NULL --Exclude approved parent
  UNION ALL

  --CHILD
  SELECT 
    c.locationid,
    c.parentid
  FROM 
    locations c
  JOIN 
    location_tree p on p.locationid = c.parentid
  LEFT JOIN
    approvelog ca ON ca.locationid = c.locationid AND ca.approved = TRUE
  WHERE
    ca.locationid IS NULL --Exclude approved children
)
SELECT 
  array_agg(locationid) 
FROM 
  location_tree

SQL小提琴：http://sqlfiddle.com/#!15/7084f/19

Answer 2

我相信@ vanlee1987的答案很好。或者，您可以将其包装在函数中，并在函数本身内执行实际递归（而不是递归SQL）。例如：

CREATE OR REPLACE FUNCTION location_tree(location_id integer)
  RETURNS integer[] AS
$BODY$
DECLARE
  result integer[];
  r locations%rowtype;
BEGIN

  select array[l.locationid]
  into result
  from
    locations l
    left join approvelog a on
      l.locationid = a.locationid
  where
    l.locationid = location_id and
   (a.approved is null or a.approved = false);

  for r in
    select
      l.locationid, l.parentid
    from
      locations l
    where
      l.parentid = any (result)
  loop
    result := array_cat(result, location_tree(r.locationid));
  end loop;

  return result;
end
$BODY$
  LANGUAGE plpgsql VOLATILE
  COST 100;

实现：

select location_tree(1);     // yields {1,2,3}
select location_tree(2);     // yields {2}
select location_tree(8);     // yields {}

如何在Postgresql中以条件递归获取记录？

2 个答案: