有限集团

Question

我有以下数据库：

date, connection_id, connection_status, file_requested
2015,     1234      ,      OK          ,file1
2015,     1234      ,      OK          ,file2
2015,     1235      ,      OK          ,file2
2014,     1236      ,      FAILURE     ,file1

我当前的查询给出了每个file_requested每个连接状态的每个日期的不同连接数，如下所示：

select date, count(distinct(connection_id)), connection_status, 
file_requested from database group by date, connection_status, file_requested;


date,     count     , connection_status, file_requested
2015,     1         ,      OK          ,file1
2015,     1         ,      OK          ,file2
2014,     1         ,      FAILURE     ,file1

但是，我还想要每个连接的每个日期的不同连接数，如下所示。 （我知道我可以在一个单独的查询中执行此操作，只需将#ad; group by file_requested＆＃34; - 但我不想要这样做。）

date,     count     , connection_status, file_requested
2015,     1         ,      OK          ,file1
2015,     1         ,      OK          ,file2
2014,     1         ,      FAILURE     ,file1
2015,     2         ,      OK          ,  --
2014,     1         ,      FAILURE     ,  --

这甚至可能吗？

Answer 1

我认为没有办法改变group by。您可以使用union：

实现您想要的效果

(select 1 as sort_column, date, count(distinct connection_id), connection_status, 
file_requested from database group by date, connection_status, file_requested)
union all
(select 2 as sort_column, date, count(distinct connection_id), connection_status, 
'--' as file_requested from database group by date, connection_status)
order by sort_column;

distinct不是函数，因此不需要括号。