我有一个类似CATEGORIES的基本表,由primary_key、parent_id、title等字段组成和一个排序整数。
我能够使用 CTE 检索结果并将它们转换为 json 数组,但我想根据排序值获取它们,除了 parent_id。
到目前为止:
with recursive parents as
(
select n.boat_type_id, n.title, '{}'::int[] as parents, 0 as level
from boat_types n
where n.parent_id is NULL
union all
select n.boat_type_id, n.title, parents || n.parent_id, level+1
from parents p
join boat_types n on n.parent_id = p.boat_type_id
where not n.boat_type_id = any(parents)
),
children as
(
select n.parent_id, json_agg(jsonb_build_object('title', n.title->>'en'))::jsonb as js
from parents tree
join boat_types n using(boat_type_id)
where level > 0 and not boat_type_id = any(parents)
group by n.parent_id
union all
select n.parent_id, jsonb_build_object('category', n.title->>'en') || jsonb_build_object('subcategories', js) as js
from children tree
join boat_types n on n.boat_type_id = tree.parent_id
)
select jsonb_agg(js) as categories
from children
where parent_id is null
上面为我提供了我想要的结果集和结构,但是我怎样才能让它们遵循节点和叶子的排序值。
示例响应:
[
{
"sorting":0,
"category":"Motor",
"subcategories":[
{
"title":"Motor Yacht",
"sorting":2
},
{
"title":"Mega Yacht",
"sorting":1
}
]
},
{
"sorting":1,
"category":"Sailing",
"subcategories":[
{
"title":"Sailing Yacht",
"sorting":2
},
{
"title":"Cruiser Racer",
"sorting":1
}
]
},
{
"sorting":2,
"category":"Catamaran",
"subcategories":[
{
"title":"Catamaran",
"sorting":2
},
{
"title":"Trimaran",
"sorting":1
}
]
},
{
"sorting":3,
"category":"Other",
"subcategories":[
{
"title":"Other",
"sorting":2
},
{
"title":"Airboat",
"sorting":1
}
]
}
]
我试过聚合 ARRAY 字段中的排序值并按它排序,但它不起作用。
最佳答案
您可以在 json_agg()
聚合中使用 order by
子句:
...
children as
(
select
n.parent_id,
json_agg(jsonb_build_object('title', n.title->>'en', 'sorting', n.sorting) order by n.sorting)::jsonb as js
from parents tree
join boat_types n using(boat_type_id)
where level > 0 and not boat_type_id = any(parents)
group by n.parent_id
union all
select
n.parent_id,
jsonb_build_object('category', n.title->>'en', 'sorting', n.sorting) || jsonb_build_object('subcategories', js) as js
from children tree
join boat_types n on n.boat_type_id = tree.parent_id
)
...
关于sql - 递归查询中的排序结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51348500/