Question

我有两个不同的查询，第一个会给我每天有拨出电话的用户数，第二个会给我每天有来电的用户数。我想要这两个查询的unioin。我的意思是我希望每天有来电或去电的用户数量。对此类结果的正确查询是什么？

1) select call_date, count(*) 
from 
(select distinct anum,call_date 
from  CDR 
where anum in (select msisdn from degree_cmp_week1)
)
group by call_date 
order by call_date;

2) select call_date, count(*) 
from 
(select distinct bnum,call_date 
from  CDR 
where bnum in (select msisdn from degree_cmp_week1)
)
group by call_date 
order by call_date;

FYI CDR是一个电话详细记录表。它包括anum（呼叫源），bnum（呼叫目的地）和呼叫日期。我正在寻找计算每天有任何活动（打电话给某人或某人称他/她）的人数。第一个查询将给我作为呼叫源的人数，第二个查询将给我作为呼叫目的地的人数。假设第一个查询给我n（A），第二个查询给我n（B）。我正在寻找n（A Union B）。如您所知n（A U B）= n（A）+ n（B） - n（交点B）。所以找到n（A交点B）对我也有帮助。

P.S：我使用的是oracle 11所以请考虑一下。

Answer 1

只需在查询中询问输入和拨出电话：

select call_date, count(num) 
from 
 (select distinct anum as num, call_date 
  from  CDR 
  where anum in (select msisdn from degree_cmp_week1)
  )
  union 
  (select distinct bnum as num ,call_date 
  from CDR
  where bnum in (select msisdn from degree_cmp_week1)
  )
group by call_date 
order by call_date;

Answer 2

这样的事情可以在一个查询中提供这两个：

with ids as (
  select distinct msisdn
  from degree_cmp_week1
), calls as (
  select
    call_date,
    nvl2 (i1.msisdn, 1, 0) as a_count,
    nvl2 (i2.msisdn, 1, 0) as b_count
  from
    CDR,
    ids i1,
    ids i2
  where
    CDR.anum = i1.msisdn (+) and
    CDR.bnum = i2.msisdn (+)
)
select
  call_date, sum (a_count) as a_count,
  sum (b_count) as b_count
from calls
group by call_date
order by call_date

SQL计数为两个不同计数的并集

2 个答案: