Question

使用Postgres 9.4。

数据：

+-----+------+
| id  | type |
+-----+------+
| 1   | A    |
+-----+------+
| 2,3 | A    |
+-----+------+
| 4   | B    |
+-----+------+

所需的输出（JSON）

[
  [{"id": "1", "type": "A"}],
  [{"id": "2", "type": "A"},{"id": "3", "type": "A"}],
  [{"id": "4", "type": "B"}]
]

我尝试过：

SELECT array_to_json(array_agg(c)) 
FROM 
(
  SELECT
    regexp_split_to_table(id, ',') AS id, 
    type 
  FROM my_table
) c;

这使我进入一个简单的json对象数组：

[
  {"id": "1", "type": "A"},
  {"id": "2", "type": "A"},
  {"id": "3", "type": "A"},
  {"id": "4", "type": "B"}]
]

如何将子查询的每个结果集（而不是每一行）包装在一个数组中？

Answer 1

我相信这可以满足您的需求

with t as (
      select *
      from (values ('1', 'A'), ('2,3', 'A'), ('4', 'B')) v(ids, type)
     )
select json_agg(c)
from (select array_to_json(array_agg( json_build_object('id', c.id, 'type', c.type))) as c
      from (select ids, regexp_split_to_table(ids, ',') AS id, type 
            from t
           ) c
      group by ids
     ) i;

Here是db <>小提琴。

Answer 2

您可以使用以下方法为单行创建JSON值：

select (select json_agg(to_json(t)) 
        from (
          select i.id, t.type 
          from unnest(string_to_array(t.ids, ',')
        ) as i(id)) t) json_id
from the_table t

您可以将其包装到派生表中，以将所有内容聚合为一个JSON值：

select json_agg(json_id)
from (
  select (select json_agg(to_json(t)) 
          from (select i.id, t.type from unnest(string_to_array(t.ids, ',')) as i(id)) t) json_id
  from the_table t
) x;

Postgres：以逗号分隔值的行到json对象的数组

2 个答案: