Question

我有这个查询，我必须计算多个字段

select 
distinct(custsegment),
(select count(distinct clid) from call_log where enteredon = '16-JUL-13') as     UniqCalls,
(select count(disconnectflag) from call_log where disconnectflag='IVR' and enteredon =     '16-JUL-13') as IvrCalls,
(select count(callerid) from call_log where enteredon = '16-JUL-13') as TotalCalls
from call_log
where enteredon = '16-JUL-13'

输出

CUSTSEGMENT  UNIQCALLS   IVRCALLS TOTALCALLS
------------ ---------- ---------- ----------
PRIORITY             12          6         12 
NORMAL               12          6         12

但问题似乎是我获得了PRIORITY和NORMAL CUSTSEGMENT的相同值，我也不确定这是否是计算它的正确方法。请建议。

Answer 1

我认为你的意思是group by。这也使您的查询更简单，因为您不需要子选择。

select 
  custsegment,
  count(distinct clid) as UniqCalls,
  count(case when disconnectflag = 'IVR' then 1 else null end) as IvrCalls,
  -- sum(case when disconnectflag = 'IVR' then 1 else 0 end) as IvrCalls,
  count('x') as TotalCalls
from call_log
where enteredon = '16-JUL-13'
group by
  custsegment

要计算IvrCalls，您可以通过几种方式进行计算。 Count计算所有非空值，因此您可以使用案例（您甚至可以省略else null）。或者您也可以使用常用的sum。

Answer 2

您的三个子查询

select count(distinct clid) from call_log where enteredon = '16-JUL-13')
select count(disconnectflag) from call_log where disconnectflag='IVR' and enteredon =     '16-JUL-13'
select count(callerid) from call_log where enteredon = '16-JUL-13')

对于结果集中的每一行，

执行完全相同。这就是为什么你看到重复的相同值。

由于你想要分组多个没有相互排斥结果的字段，我会说（这可能不是最好的方法）组对你感兴趣的每个项目，然后结合你的结果。

Answer 3

试试这个：

Select 
c.custsegment,
count(distinct clid) as     UniqCalls,
count(callerid) as TotalCalls
from call_log c 
inner join
(select count(disconnectflag) as IvrCalls, custsegment 
from call_log where disconnectflag='IVR' and enteredon = '16-JUL-13' group by custsegment) t

on c.custsegment=t.custsegment

where enteredon = '16-JUL-13'
group by c.custsegment

单个查询中的多个计数

3 个答案: