我有以下情况。每一行都有一个写在表上的时间戳。现在我想每天评估在早上5点之前插入了多少行以及之后插入了多少行。怎么可以做到?
答案 0 :(得分:2)
您可以使用HH24格式获取24小时制的小时:
select trunc(created_Date) as the_day
,sum(case when to_number(to_char(created_Date,'HH24')) < 5 then 1 else 0 end) as before_five
,sum(case when to_number(to_char(created_Date,'HH24')) >= 5 then 1 else 0 end) as after_five
from yourtable
group by trunc(created_Date)
根据用户对5:10的评论,显示5之前和之后的时间戳:
select trunc(created_Date) as the_day
,sum(case when to_number(to_char(created_Date,'HH24')) < 5 then 1 else 0 end) as before_five
,sum(case when to_number(to_char(created_Date,'HH24')) >= 5 then 1 else 0 end) as after_five
from (
-- one row januar 1 just after 5:00 a.m.
select to_Date('01/01/2015 05:10:12','dd/mm/yyyy hh24:mi:ss') as created_date from dual
union all
-- one row Januar 2 just before 5:00 a.m.
select to_Date('02/01/2015 04:59:12','dd/mm/yyyy hh24:mi:ss') as created_date from dual
)
group by trunc(created_Date);
THE_DAY, BEFORE_FIVE, AFTER_FIVE
02/01/2015, 1, 0
01/01/2015, 0, 1
答案 1 :(得分:0)
假设您的时间戳是DATE列:
select trunc(date_written) as day
, count (case when (date_written-trunc(date_written))*24 < 5 then 1 end) before_5_count
, count (case when (date_written-trunc(date_written))*24 >= 5 then 1 end) after_5_count
from mytable
group by trunc(date_written)
答案 2 :(得分:0)
select to_char(time_column, 'dd/mm/yyyy'),
sum( decode ( greatest(extract(hour from time_column), 5), extract(hour from time_column), 1, 0)) post_5,
sum( decode ( greatest(extract(hour from time_column), 5), extract(hour from time_column), 0, 1)) pre_5
from test_time
group by to_char(time_column, 'dd/mm/yyyy')