Counting Records based on Time

Question

Hi,
This is my Records
Date Rolls
09-oct-12 08:00:00 10
09-oct-12 08:00:50 20
10-oct-12 07:00:00 5
10-oct-12 10:00:00 10
Morning 8 to next morning 8 o'lcock to be taken as single day.
Output:
09-oct-12 35 (10+20+5)
10-oct-12 10
Pl. help How to achieve this.
Thanks
Kavi

Answer 1

When you wish a "skewed offset" definition of a day, you can subtract the required number of hours and then trunc to whole days.
Like this:
SQL> with data as (
  2     select to_date('09-10-2012 08:00:00','DD-MM-YYYY HH24:MI:SS') thedate, 10 rolls from dual union all
  3     select to_date('09-10-2012 08:00:50','DD-MM-YYYY HH24:MI:SS') thedate, 20 rolls from dual union all
  4     select to_date('10-10-2012 07:00:00','DD-MM-YYYY HH24:MI:SS') thedate,  5 rolls from dual union all
  5     select to_date('10-10-2012 10:00:00','DD-MM-YYYY HH24:MI:SS') thedate, 10 rolls from dual
  6  )
  7  select to_char( trunc(data.thedate - interval '8' hour)
  8                , 'DD-MM-YYYY HH24:MI:SS' ) theday
  9       , sum(rolls) rolls
10    from data
11   group by trunc(data.thedate - interval '8' hour)
12   order by theday
13  /
THEDAY                   ROLLS
09-10-2012 00:00:00         35
10-10-2012 00:00:00         10