Question

第一次查询：

SELECT date_trunc('day', date1) as date, COUNT(*) AS count_a 
FROM table_a GROUP BY date

结果：

date                    count_a
2014-04-01 00:00:00.0   1011642
2014-04-02 00:00:00.0   309048

第二次查询：

SELECT date_trunc('day', date1) as date, COUNT(*) AS count_b 
FROM table_b GROUP BY date

结果：

date                    count_b
2014-04-01 00:00:00.0   2342
2014-04-02 00:00:00.0   43432

我想得到：

date                    count_a  count_b
2014-04-01 00:00:00.0   1011642  2342
2014-04-02 00:00:00.0   309048   43432

这可能吗？

Answer 1

简单的UNION ALL应该可以实现。

SELECT date1 date, SUM(a) count_a, SUM(b) count_b 
FROM (
  SELECT DATE_TRUNC('day', date1) date1, 1 a, 0 b FROM table_a 
  UNION ALL 
  SELECT DATE_TRUNC('day', date1) date1, 0 a, 1 b FROM table_b
) z 
GROUP BY date;

编辑：Here's an SQLfiddle。

Answer 2

使用FULL [OUTER] JOIN：

SELECT *
FROM  (
   SELECT date1::date AS day, COUNT(*) AS count_a
   FROM   table_a
   GROUP  BY 1
   ) a
FULL JOIN (
   SELECT date1::date AS day, COUNT(*) AS count_b
   FROM   table_b
   GROUP  BY 1
   ) b USING (day);

USING子句自动折叠到结果中的一个列day。
要将日期投放到date。更快，更简单。

SQLfiddle（基于约阿希姆）。

次要差异：此查询返回NULL天数，一侧没有任何行。使用 COALESCE(count_b, 0) 来获取0。

PostgreSQL在每个查询中合并两个查询，其中COUNT和GROUP BY

2 个答案: