PostgreSQL 合并 COUNT 和 GROUP BY 结果答案

【问题标题】：PostgreSQL merge COUNT and GROUP BY resultPostgreSQL 合并 COUNT 和 GROUP BY 结果
【发布时间】：2018-05-23 08:37:07
【问题描述】：

select type, created_at::date , count(type)
from txns
group by created_at::date, type
order by created_at::date;

上面的 SQL 查询给出以下输出：

  type   | created_at | count 
---------+------------+-------
 full    | 2017-05-20 |     2
 virtual | 2017-05-20 |     2
 full    | 2017-05-21 |     1
 virtual | 2017-05-21 |     1
 full    | 2017-05-22 |     3

如何将上述结果按created_at 分组以获得以下输出：

 created_at | full_count | virtual_count
------------+-----------------
 2017-05-20 |     2     |  2
 2017-05-21 |     1     |  1
 2017-05-22 |     3     |  0

我想在一行中按created_at 获取full 和virtual 类型计数。

【问题讨论】：

标签： sql postgresql postgresql-9.5

【解决方案1】：

这应该可行：

with so as (
select type, created_at::date , count(type) from txns group by created_at::date, type order by created_at::date
)
 select
  created_at
, sum(case when type = 'full' then count else 0 end) full_count
, sum(case when type = 'virtual' then count else 0 end) virtual_count
from so
group by created_at;

 created_at | full_count | virtual_count
------------+------------+---------------
 2017-05-20 |          2 |             2
 2017-05-21 |          1 |             1
 2017-05-22 |          3 |             0
(3 rows)

【讨论】：

你太棒了，谢谢！等待其他结果，否则我会接受它非常感谢。

【解决方案2】：

我会在没有 CTE/子查询的情况下这样做：

select created_at::date,
       sum( (type = 'Full')::int ) as cnt_full,
       sum( (type = 'Virtual')::int ) as cnt_virtual
from txns
group by created_at::date
order by created_at::date;

Postgres 经常实现 CTE，因此这甚至可能具有轻微的性能优势。

注意：您还可以在更新的 Postgres 版本中使用 filter：

select created_at::date,
       count(*) filter (where type = 'Full') as cnt_full,
       count(*) filter (where type = 'Virtual') as cnt_Virtual
from txns
group by created_at::date
order by created_at::date;

【讨论】：

这个方案比较好，“type”不应该在组里。

【解决方案3】：

如果你有两种以上的类型，在进一步处理中你可以处理 hstore：

WITH q AS ( SELECT type, created_at::date , count(type) FROM txns GROUP BY 1,2)
SELECT created_at, hstore(array_agg(type), array_agg(count::text))
FROM q GROUP BY 1;

【讨论】：