Grouping records hour by hour or day by day and filling gaps with zero or null(按小时或按天对记录进行分组,并用零或空值填充空白)
本文介绍了按小时或按天对记录进行分组,并用零或空值填充空白的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我写了一个查询,按小时计算记录:
I have written a query that counts records hour by hour:
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
我也有这些按天和月分组的查询!
Also I have these queries that group by day and month too!
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');
我也需要用零或空来填充它们的间隙.任何帮助都非常感谢.
I need their gaps to be filled with zero or null too. Any help is really appreciated.
推荐答案
尝试:
第一个查询(按小时):
try:
first query (by hour):
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(按天):
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');
第三次查询(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');
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本文标题为:按小时或按天对记录进行分组,并用零或空值填充空白
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