为什么 json_agg() 会抱怨子查询返回多于一行？

Question

我想选择每个宠物品种的数量，并按其主人和类型分组。

我编写了一个子查询，它按宠物类型返回品种计数，这几乎是我所需要的，只是我需要 json 数组中的结果。

按类型分类
{“type”：“CAT”，“breed_count”：[{“孟加拉”：1，“缅甸”：2}]}
{"type": "DUG", "breed_count": [{"GERMAN_SHEPHERD": 1, "AKITA": 2}]}

我预计 json_agg() 函数会执行此操作，但它返回一个错误：[21000] ERROR: more more row returned by a subquery using as an expression.

这是我的完整查询：

SELECT 
  o.id, 
  JSON_AGG(
    (
      SELECT 
        JSONB_BUILD_OBJECT(
          'type', 
          p1.type, 
          'breed_count', 
          JSONB_BUILD_ARRAY(
            JSONB_OBJECT_AGG(
              p1.breed, 
              (
                SELECT 
                  COUNT(p2.id) 
                FROM 
                  pet p2 
                WHERE 
                  p2.owner_id = p1.owner_id 
                  AND p2.breed = p1.breed 
                  AND p2.type = p1.type
              )
            )
          )
        ) AS breed_by_type 
      FROM 
        pet p1 
      GROUP BY 
        p1.type
    )
  ) 
FROM 
  owner o 
  JOIN pet p ON p.owner_id = o.id 
GROUP BY 
  o.id;

如何将子查询的结果收集到json数组中？

示例：d-fiddle

Answer 1

这是另一种更优化的选择：

select jsonb_build_object(
            'type',
            type,
            'breed_count',
            json_agg(jsonb_build_object(breed, cnt))
       ) as breed_by_type
            
from (
  select type, breed, count(*) as cnt
  from pet
  group by type, breed
) as s
group by type;

结果：

breed_by_type
--------------
{"type":"CAT","breed_count":[{"BURMESE":2},{"BENGAl":1}]}
{"type":"DOG","breed_count":[{"AKITA":2},{"GERMAN_SHEPHERD":1}]}