获取 Postgres NOSQL json 数据中 json 数组的总和答案

【问题标题】：Get aggregate sum of json array in Postgres NOSQL json data获取 Postgres NOSQL json 数据中 json 数组的总和
【发布时间】：2021-03-24 01:05:47
【问题描述】：

如何在 postgres json 选择中从“refunds”数组中获取聚合 SUM(amount) 以下是我的数据架构和结构：

表名：transactions

列名：data

{
  "id": "tran_6ac25129951962e99f28fa488993",
  "amount": 1200,
  "origin_amount": 3900,
  "status": "partial_refunded",
  "description": "Subscription#sub_a67d59efb2bcbf73485a ",
  "livemode": false,
  "refunds": [
    {
      "id": "refund_ee4192ffb6d2caa490a1",
      "amount": 1200,
      "status": "refunded",
      "created_at": 1426412340,
      "updated_at": 1426412340,
    },
    {
      "id": "refund_0e4a34e4ee7281d369df",
      "amount": 1500,
      "status": "refunded",
      "created_at": 1426412353,
      "updated_at": 1426412353,
    }
  ]
}

输出应该是：1200+1500 = 2700

Output
|---------
|total
|---------
|2700

请提供全局解决方案，不要提供静态数据

【问题讨论】：

标签： json postgresql nosql

【解决方案1】：

这应该适用于 9.3+

WITH x AS( SELECT
'{
  "id": "tran_6ac25129951962e99f28fa488993",
  "amount": 1200,
  "origin_amount": 3900,
  "status": "partial_refunded",
  "description": "Subscription#sub_a67d59efb2bcbf73485a ",
  "livemode": false,
  "refunds": [
    {
      "id": "refund_ee4192ffb6d2caa490a1",
      "amount": 1200,
      "status": "refunded",
      "created_at": 1426412340,
      "updated_at": 1426412340
    },
    {
      "id": "refund_0e4a34e4ee7281d369df",
      "amount": 1500,
      "status": "refunded",
      "created_at": 1426412353,
      "updated_at": 1426412353
    }
  ]
}'::json as y),
refunds AS(
SELECT json_array_elements(y->'refunds') as j FROM x)
SELECT sum((j->>'amount')::int) FROM refunds;

【讨论】：

这里 json 数据是可变的，所以我可以使用“数据”表示字段名称而不是固定字符串。我是 postgres 的新手，你能提供精确的查询列“数据”和动态，而不是静态
refunds CTE 从 x.y 中选择，如果你想从 tbl1.data 中选择就这样做WITH refunds AS( SELECT json_array_elements(data->'refunds') as j FROM tbl1) (...)
我跑了WITH x AS( SELECT json_array_elements(data->'refunds') as y FROM transactions ), refunds AS( SELECT json_array_elements(y->'refunds') as j FROM x) SELECT sum((j->>'amount')::int) FROM refunds;并得到了错误ERROR: cannot call json_array_elements on a scalar如果我错了请告诉我
@mukund 看来您的数据并不总是像示例中那样。如果您使用的是 9.4，请尝试 SELECT json_array_elements(y->'refunds') as j FROM x WHERE son_typeof(y->'refunds') = 'array。

【解决方案2】：

WITH AllRefunds AS ( SELECT jsonb_array_elements(data->'refunds') AS refund FROM transactions)
SELECT SUM( CAST ( refund ->> 'amount' AS INTEGER )) FROM AllRefunds;

如果您需要知道查询是如何构建的：

WITH AllRefunds AS ( SELECT jsonb_array_elements(data->'refunds') FROM transactions)
SELECT * FROM AllRefunds;

这将从交易表中找到的数组refunds 中选择所有元素作为JSON 对象（通过-> 完成）并将其存储在新表AllRefunds 中。这个新表只包含一个未命名的列。

WITH AllRefunds AS ( SELECT jsonb_array_elements(data->'refunds') AS refund FROM transactions)
SELECT * FROM AllRefunds;

这里添加的（第二个）AS 将AllRefunds 中当前未命名的列重命名为refund

WITH AllRefunds AS ( SELECT jsonb_array_elements(data->'refunds') AS refund FROM transactions)
SELECT SUM( CAST ( refund ->> 'amount' AS INTEGER )) FROM AllRefunds;

我们的数组条目是 JSON 对象。因此，我们将字段 amount 作为带有 ->> 的简单字符串返回，然后我们将其转换为 Integers 和 SUM 所有条目。

【讨论】：