Question

我正在尝试使用以下查询来根据时间获取行。当我得到行时，我想平均列中的值，并返回一行，其平均值为JSON。希望它能用一些代码来理解。这是我开始的PostgreSQL查询：

select row_to_json(row)
  from (
    SELECT "responses"."index", "responses"."created_at",
    ROUND(AVG("responses"."value")) AS value
    FROM "responses"
    WHERE "responses"."time" = '1416177080'
    GROUP BY "responses"."index", "responses"."created_at"
  ) row

产生如下输出：

{"index":4,"created_at":"2014-11-16 22:31:27","value":5}
{"index":4,"created_at":"2014-11-16 22:31:23","value":-5}
{"index":4,"created_at":"2014-11-16 22:31:24","value":0}
{"index":4,"created_at":"2014-11-16 22:31:25","value":-1}
{"index":4,"created_at":"2014-11-16 22:31:26","value":1}
(5 rows)

当我真正想要的只是一行时，平均值为值。像这样：

{"index_of_affiliation":4,"created_at":"2014-11-16 22:31:27","value":0}

我觉得created_at可能会让我失望，但我想在JSON中返回一个值。任何关于如何获得我想要的平均值的行的反馈都会有所帮助。

Answer 1

这些方面的某些内容可能有用。

select row_to_json(row)
from (select "index" index_of_affiliation, 
             max("created_at") max_created_at, 
             round(avg("value")) avg_value
      from responses
      where "time" = '1416177080'
      group by "index"
     ) row ;

从长远来看，你最好使用更具描述性的名字而不是＆＃34; index＆＃34;，＆＃34; value＆＃34;，＆＃34; time＆＃34;。

PostgreSQL：获取与列匹配的行的平均值

1 个答案: