Question

我编写了一个按小时统计记录的查询：

select TO_CHAR(copied_timestamp, 'YYYY-MM-DD HH24'),count(*) from req group by
TO_CHAR(copied_timestamp, 'YYYY-MM-DD HH24');

结果是：

2012-02-22 13    2280
2012-02-22 15    1250
2012-02-22 16    1245
2012-02-22 19    1258

但我需要这样的结果：

2012-02-22 13    2280
2012-02-22 14    0
2012-02-22 15    1250
2012-02-22 16    1245
2012-02-22 17    0
2012-02-22 18    0
2012-02-22 19    1258

此外，我也有按天和月分组的查询！

select TO_CHAR(copied_timestamp, 'YYYY-MM-DD'),count(*)  from req
group by TO_CHAR(copied_timestamp, 'YYYY-MM-DD');

select TO_CHAR(copied_timestamp, 'YYYY-MM'),count(*)  from req
group by TO_CHAR(copied_timestamp, 'YYYY-MM');

我需要将他们的差距填充为零或零。任何帮助都非常感谢。

Answer 1

尝试：
第一次查询（按小时）：

with t as (
  select mnd + ((level-1)/24) ddd
  from
  (select trunc(min(copied_timestamp),'hh') mnd, trunc(max(copied_timestamp),'hh') mxd from req) v
  connect by mnd + ((level-1)/24) <= mxd
  )
select to_char(trunc(d1, 'hh'), 'yyyy-mm-dd hh24'), count(d2) from 
(select nvl(copied_timestamp, ddd) d1, copied_timestamp d2 from req right outer join (
  select ddd from t) ad on ddd = trunc(copied_timestamp, 'hh'))
group by trunc(d1, 'hh');

第二次查询（按日）：

with t as (
      select mnd + level-1 ddd
      from
      (select trunc(min(copied_timestamp),'dd') mnd, trunc(max(copied_timestamp),'dd') mxd from req) v
      connect by mnd + level-1 <= mxd
      )
    select to_char(trunc(d1, 'dd'), 'yyyy-mm-dd'), count(d2) from 
    (select nvl(copied_timestamp, ddd) d1, copied_timestamp d2 from req right outer join (
      select ddd from t) ad on ddd = trunc(copied_timestamp, 'dd'))
    group by trunc(d1, 'dd');

第三个查询（按月）：

with t as (
      select add_months(mnd, level-1) ddd
      from
      (select trunc(min(copied_timestamp),'mm') mnd, trunc(max(copied_timestamp),'mm') mxd from req) v
      connect by add_months(mnd, level-1) <= mxd
      )
    select to_char(trunc(d1, 'mm'), 'yyyy-mm'), count(d2) from 
    (select nvl(copied_timestamp, ddd) d1, copied_timestamp d2 from req right outer join (
      select ddd from t) ad on ddd = trunc(copied_timestamp, 'mm'))
    group by trunc(d1, 'mm');

将记录按小时或按日分组，并用零或空填充缺口

1 个答案: