在时间之前计算行数

Question

我有以下情况。每行在 table 上写入时都有一个时间戳。现在我想评估每天早上 5 点之前插入了多少行，之后插入了多少行。怎么做到的？？

Answer 1

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)

Answer 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 前后的时间戳：

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

Answer 3

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')

在时间之前计算行数

count rows before time

sql

oracle

evaluate