postgresql - postgres复杂查询

Question

我不知道是否可以进行这样的查询。问题是我有一张表，其中有一些日期数字。假设我有 3 列:日期、值(value)、好/坏

即:

2014-03-03 100 Good
2014-03-03 15 Bad
2014-03-04 120 Good
2014-03-04 10 Bad

我想选择并减去 Good-Bad:

2014-03-03 85
2014-03-04 110

这可能吗？我想了很多，还没有主意。如果我在单独的表中有 Good 和 Bad 值，那将相当简单。

Answer 1

诀窍是将您的表重新连接到它自己，如下所示。 myTable as A 将只读取Good 行，myTable as B 将只读取Bad 行。然后根据日期将这些行连接成一个单行。

SQL Fiddle Demo

select 
a.date
,a.count as Good_count
,b.count as bad_count
,a.count-b.count as diff_count
from myTable as a
inner join myTable as b
on a.date = b.date and b.type = 'Bad'
where a.type = 'Good'

返回的输出:

DATE                            GOOD_COUNT  BAD_COUNT   DIFF_COUNT
March, 03 2014 00:00:00+0000    100           15         85
March, 04 2014 00:00:00+0000    120           10         110

另一种方法是使用 Group by 而不是 inner join:

select 
a.date
,sum(case when type = 'Good' then a.count else  0 end) as Good_count
,sum(case when type = 'Bad' then a.count else  0 end) as Bad_count
,sum(case when type = 'Good' then a.count else  0 end)  - 
    sum(case when type = 'Bad' then a.count else  0 end) as Diff_count
from myTable as a
group by a.date
order by a.date

两种方法产生相同的结果。