Sql 有没有办法将JSON对象数组与其他视图/表中的另一个JSON对象数组合并？

因此，有两个表我想按年龄分组，这样我就可以从两个表中找到该年龄组的总人数。到目前为止，我能够做到这一点，但我希望JSON数据的方式是错误的

这是我的方法：

SELECT temp.age, sum(temp.total), json_agg(temp.data_recs) as data_recs FROM (

        SELECT  
            '31-40' as age, COUNT(*) as total, 
            json_agg(
                  CASE
                  WHEN table_1.age BETWEEN 31 AND 40 
                  THEN json_build_object(
               'country', table_1.country,
               'Name', table_1.name
            )
            ELSE json_build_object(
                  'country', '',
                  'Name', ''
                )
                END
                  ) AS data_recs 
            from table_1 
        WHERE table_1.age BETWEEN 31 AND 40

        UNION ALL

        SELECT  
            '31-40' as age, COUNT(*) as total, 
            json_agg(
                  CASE
                  WHEN table_2.age BETWEEN 31 AND 40
                  THEN json_build_object(
               'country', table_2.country,
               'Name', table_2.name
            )
            ELSE json_build_object(
                  'country', '',
                  'Name', ''
                )
                END
                  ) AS data_recs 
            from table_2 
        WHERE table_2.age BETWEEN 31 AND 40

  ) AS temp group by temp.age 

第1表中的结果如

  age  |  total |  data_recs
'31-40'|   2    | [{"name":"John","country":"USA"},{"name":"Eddy","country":"USA"}]

  age  |   total |  data_recs
'31-40'|   2     | [{"name":"Cienna","country":"Italy"},{"name":"Rosie","country":"Italy"}]

第二个表eg的结果

  age  |  total |  data_recs
'31-40'|   2    | [{"name":"John","country":"USA"},{"name":"Eddy","country":"USA"}]

  age  |   total |  data_recs
'31-40'|   2     | [{"name":"Cienna","country":"Italy"},{"name":"Rosie","country":"Italy"}]

到目前为止，我能够实现

  age  |   total |  data_recs
'31-40'|   4     | [[{"name":"John","country":"USA"},{"name":"Eddy","country":"USA"}],[{"name":"Cienna","country":"Italy"},{"name":"Rosie","country":"Italy"}]]

JSON数组1=

[{“name”：“John”，“country”：“USA”}，{“name”：“Eddy”，“country”：“USA”}]

JSON数组2=

[{“name”：“ciena”，“country”：“Italy”}，{“name”：“Rosie”，“country”：“Italy”}]

我想要的是：-

     age  |   total |  data_recs
  '31-40' |   4     | [{"name":"John","country":"USA"},{"name":"Eddy","country":"USA"},{"name":"Cienna","country":"Italy"},{"name":"Rosie","country":"Italy"}]

我想要的组合JSON应该是这样的：

[{"name":"John","country":"USA"},{"name":"Eddy","country":"USA"},{"name":"Cienna","country":"Italy"},{"name":"Rosie","country":"Italy"}]

---

有没有办法做到这一点或其他方法？

基本上，您的代码如下所示

SELECT
    json_agg(json_objects from UNION)
FROM (
    SELECT
        json_agg(
            json_build_object()
        )
    FROM
        ...

    UNION

    SELECT
        json_agg(
            json_build_object()
        )
    FROM
        ...
)

因此，首先创建数组，然后聚合数组。但是如果您省略了第一步，那么高级的

json\u agg（）

将聚合您使用

json\u object\u build（）创建的json对象

因此，您的解决方案应该只有一个聚合：

SELECT
    json_agg(json_arrays from UNION)
FROM (
    SELECT
        json_build_object()
    FROM
        ...

    UNION

    SELECT
        json_build_object()
    FROM
        ...
)

有没有可能在不从内部删除json_agg（）的情况下得到结果，然后使用一些最不常用的函数呢？我看不出有什么方法是有意义的。当然，您可以使用内部json_agg（）并在之后使用json_array_elements（）取消对元素的测试。但为什么要这样做：聚合只是为了在。。。这可能是代码：