在Json对象中使用JSON1和SQLite3执行聚合

我刚开始使用SQLite 3和JSON1支持

我已经创建了一个包含多个属性的数据库。 其中一个属性是json对象

我要做的是在这个对象中执行聚合

正在运行以下查询：

select json_extract(report, '$.some_attribute')
from table1
group by var1
having date == max(date);

返回以下内容：

[{"count":55,"label":"A"},{"count":99,"label":"B"}, {"count":1,"label":"C"}]
[{"count":29,"label":"A"},{"count":285,"label":"B"},{"count":461,"label":"C"}]
[{"count":6642,"label":"A"},{"count":24859,"label":"B"},{"count":3031,"label":"C"}]
[{"count":489,"label":"A"},{"count":250,"label":"B"},{"count":74,"label":"C"}]

现在，我想做的是按标签键分组，例如，求和计数键

输出应如下所示：

[{"label": A, 'count': 7215},
 {"label": B, 'count': 25493},
 {"label": C, 'count': 3567}]

select sum(A) as A, sum(B) as B, sum(C) as C
from (
select json_extract(report,
                '$.some_attribute[0].count') as A,
       json_extract(report,
                '$.some_attribute[1].count') as B,
       json_extract(report,
                '$.some_attribute[0].count') as C
from table1
group by var1
having date == max(date));

或者这个：

  A,       B,     C
7215,    25493,  3567

我试着实现后一种方法，如下所示：

[{"label": A, 'count': 7215},
 {"label": B, 'count': 25493},
 {"label": C, 'count': 3567}]

select sum(A) as A, sum(B) as B, sum(C) as C
from (
select json_extract(report,
                '$.some_attribute[0].count') as A,
       json_extract(report,
                '$.some_attribute[1].count') as B,
       json_extract(report,
                '$.some_attribute[0].count') as C
from table1
group by var1
having date == max(date));

问题是，如何确保数组中的所有对象都以相同的方式排序。所以这可能会引起问题

有什么解决办法吗？谢谢

如果您“取消嵌套”从第一个

json\u extract返回的json字符串，它将变得微不足道。这在我的报告中起了作用：

WITH result as (select jsonString from jsonPlay)
select json_extract(value,'$.label') label,SUM(json_extract(value,'$.count'))
from result,json_each(jsonString) 
group by label

给出这个结果：

A| 7215
 B| 25493
 C| 3567