PostgreSQL 合并 COUNT 和 GROUP BY 结果

Question

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

上面的 SQL 查询给出了以下输出：

  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 计算 full 和 virtual 类型。

Answer 1

这应该有效：

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)

Answer 2

如果你有两个以上的类型，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;

Answer 3

我会在没有 CTE/subquery 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 经常具体化 CTE，因此这甚至可能具有轻微的性能优势。

注意：您还可以在较新版本的 Postgres 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;

PostgreSQL 合并 COUNT 和 GROUP BY 结果

PostgreSQL merge COUNT and GROUP BY result

sql

postgresql

postgresql-9.5