Question

我有一堆动态调查结果存储在一个带有JSON列的表中，如下所示：

{"regional":"never","local":"sometimes","personal":"often","bike":"field_4","walk":"field_5","carpool":"field_4","carshare":"often","rideshare":"sometimes"}

键是整个数据集中动态存在的值。我需要找到一种有效的方法来聚合密钥并计算相关结果。该数据代表了李克特量表调查问题，即

＆＃34;你多久去一个公交车站？不经常，有时，经常＆＃34;

我想生成的结果数据集是：

{
    "regional": { "sometimes": 20, "often": 10, "never": 5 },
    "bike": { "sometimes": 20, "often": 10, "never": 5 }
    "walk": { "sometimes": 20, "often": 10, "never": 5 }
}

我已尝试将JSON_EACH分解为键/值对，但我现在无法汇总密钥和潜在结果的频率。这就是我到目前为止所做的：

  SELECT
    json_data.key,
    json_data.value
  FROM
    "SubmissionData" AS sd,
    JSON_EACH_TEXT(sd.data) AS json_data
  WHERE
    sd.key = 'transit'

虽然在这里肯定遇到了障碍。

感谢。

Answer 1

使用json_object_agg()，例如：

with my_table(data) as (
values
    ('{"regional":"never","local":"sometimes","personal":"often"}'::json),
    ('{"regional":"often","local":"sometimes","personal":"often"}'::json)
)

select json_object_agg(key, vals)
from (
    select key, json_object_agg(value, count) vals
    from (
        select key, value, count(*)
        from
            my_table,
            json_each_text(data)
        group by 1, 2
        ) s
    group by 1
    ) s;

                                              json_object_agg                                               
------------------------------------------------------------------------------------------------------------
 { "regional" : { "never" : 1, "often" : 1 }, "personal" : { "often" : 2 }, "local" : { "sometimes" : 2 } }
(1 row)

按键

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