Oracle PLSQL truncate datetime to specific hours

Question

I have an Oracle PLSQL code generating a list of datetime stamps and I would like to truncate them to the specific hours of 7am and 7pm rather than the beginning of the day.

For example:

• 01/03/2013 0700 becomes 01/03/2013 0700
• 01/03/2013 1235 becomes 01/03/2013 0700
• 01/03/2013 1932 becomes 01/03/2013 1900
• 02/03/2013 0612 becomes 01/03/2013 1900

My code is currently:

 SELECT  TRUNC(TRUNC(SYSDATE,'hh') + 1/24 - (ROWNUM) / 24, 'dd')  as shift_date
 FROM widsys.times
 ORDER BY SYSDATE

Thanks

Answer 1

Without conditionals :)

Select your_date, 
  trunc(your_date - 7/24) +                       --the date
  trunc(to_char(your_date - 7/24,'hh24')/12)/2 +  --wich half of day
  7/24                                            --shift the hour
from 
your_table;

See a fiddle.

Answer 2

with data(time) as (
  select to_date('2013-09-19 00:00:00', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 06:45:44', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 08:12:25', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 18:59:59', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 19:00:00', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 20:15:35', 'YYYY-MM-DD HH24:MI:SS') from dual union all
  select to_date('2013-09-19 23:59:59', 'YYYY-MM-DD HH24:MI:SS') from dual
)
select d.time,
case
  when to_number(to_char(d.time, 'HH24')) >= 19 then
    trunc(d.time) + 19/24
  when to_number(to_char(d.time, 'HH24')) >= 7 then
    trunc(d.time) + 7/24
  else
    trunc(d.time - 1) + 19/24
end as shift_date
from data d
;