Question

如果我可以从表格栏中执行此查询：

SELECT count(distinct bar_id) as count, 
       date_trunc('day', created_at) as date
FROM 
  bars
GROUP BY 
  date 
ORDER BY 
  date desc

来自表foos的这个查询：

SELECT count(distinct foo_id),
       date_trunc('day', created_at) as date
FROM 
  foos
GROUP BY 
  date 
ORDER BY 
  date desc

我得到了很好的清单，如：

22, 2017-09-14
31, 2017-09-13
18, 2017-09-12

和此：

34, 2017-09-14
55, 2017-09-13
67, 2017-09-12

在一个查询中，如何将这两者结合起来产生：

22/34, 0.64, 2017-09-14
31/55, 0.56, 2017-09-13
18/67, 0.26, 2017-09-12

即。我想将table1中的每个计数除以按日分组的table2中的日期。 22/34 = 0.64,31 / 55 = 0.56等。

Answer 1

你可以加入他们：

SELECT bars.counts * 1.0 / foos.counts
    , bars.date
FROM
(
SELECT count(distinct bar_id) as counts, 
       date_trunc('day', created_at) as date
FROM 
  bars
GROUP BY 
  date 
) bars_q
JOIN
(
SELECT count(distinct foo_id) as counts,
       date_trunc('day', created_at) as date
FROM 
  foos
GROUP BY 
  date 
) foos_q
ON bars_q.date = foos_q.date

Answer 2

您可以使用FULL OUTER JOIN检索两个表上的记录，还可以显示没有匹配日期的记录

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在Postgres中，如何通过查询组合两个组并对结果进行简单的划分？

2 个答案: