CODEWITHSUNDEEP
sqloracle-databasedatetime
MontyPython
MontyPython

Reputation: 2994

How to eliminate non-working hours in Oracle

I have two columns of DateTime type. The first one stores the DateTime when a process it started, the other one stores the DateTime when that process is finished. I want to calculate the total working hours put in to complete the work. The working hours are 10 am to 7 pm and Sundays are off.

Here are a few snippets that I've jotted down:

       SELECT col1, col2, floor(((date2-date1)*24*60*60)/3600)
       || ' HOURS ' ||
       floor((((date2-date1)*24*60*60) -
       floor(((date2-date1)*24*60*60)/3600)*3600)/60)
       || ' MINUTES ' ||
       round((((date2-date1)*24*60*60) -
       Floor(((date2-date1)*24*60*60)/3600)*3600 -
       (floor((((date2-date1)*24*60*60) -
       floor(((date2-date1)*24*60*60)/3600)*3600)/60)*60) ))
       || ' SECS ' Time_Difference
       From Table_Name;

And, 

       Select To_Number(To_Char(date1, 'HH24')) || ':'
       ||   to_number(to_char(date1, 'MI')) || ':'||
       to_number(to_char(date1, 'SS')) from Table_Name

Please help.

To the above solution you have mentioned, this is the result!

START DATE          DAY     FINISH DATE     DAY  Date Diff   Total Hours   Work Hours 

07-AUG-12 21:55:21  TUE 08-AUG-12 11:09:10  WED 0 13:13:49.0        13   13 
13-NOV-12 15:45:25  TUE 14-NOV-12 10:41:42  WED 0 18:56:17.0        18   18 
20-DEC-12 20:31:03  THU 21-DEC-12 11:03:36  FRI 0 14:32:33.0        14   14 
14-MAR-13 20:39:00  THU 15-MAR-13 11:00:04  FRI 0 14:21:4.0         14   14 
07-JUN-12 21:17:36  THU 08-JUN-12 11:02:23  FRI 0 13:44:47.0        13   13 
18-SEP-12 20:48:27  TUE 19-SEP-12 11:07:35  WED 0 14:19:8.0         14   14

New Scenario: https://stackoverflow.com/questions/17248741/how-to-get-business-hours-between-two-dates-in-oracle

Upvotes: 0

Views: 3739

Answers (2)

Noel
Noel

Reputation: 10525

If i understand correctly, you want to calculate the difference between the start and finish date excluding the time before 10 am and after 7 pm.

Here's the sample query and sql fiddle.

SELECT start_time,
       finish_time,
       interval_time,
       EXTRACT (HOUR FROM interval_time), --extract the hours,mins and seconds from the interval
       EXTRACT (MINUTE FROM interval_time),
       EXTRACT (SECOND FROM interval_time)
  FROM (SELECT start_time,
               finish_time,
               NUMTODSINTERVAL (
                    CASE
                       WHEN finish_time - TRUNC (finish_time) > (19 / 24) --if finish time is after 7pm
                       THEN
                          TRUNC (finish_time) + (19 / 24)      --set it to 7pm
                       ELSE
                          finish_time      --else set it to actual finish time
                    END
                  - CASE
                       WHEN start_time - TRUNC (start_time) < (10 / 24) --if start time is before 10 am
                       THEN
                          TRUNC (start_time) + (10 / 24)    --set it to 10 am.
                       ELSE
                          start_time    --else set it to the actual start time
                    END,
                  'day') --subtract the both and convert the resulting day to interval
                  interval_time
          FROM timings);

What I have done is,

  • Check if the start time is before 10 am and finish time is after 7 pm. If so, set the time to 10 am and 7 pm.
  • Then subtract the dates and convert the resulting days to Interval Type.
  • Then extract the hours, mins and seconds from the Interval.

Note: This query assumes that both dates fall on same day and both are not before 10am or after 7 pm.

UPDATE: To exclude holidays, the query will become complicated. I suggest writing three functions and use these functions in the query.

1st function:

FUNCTION modify_start_time (p_in_dte DATE) RETURN DATE
----------------------------------
IF p_in_dte - TRUNC (p_in_dte) < (10 / 24)
THEN
   RETURN TRUNC (p_in_dte) + (10 / 24);
ELSIF p_in_dte - TRUNC (p_in_dte) > (19 / 24)
THEN
   RETURN TRUNC (p_in_dte) + 1 + (10 / 24);
ELSE
   RETURN p_in_dte;
END IF;

If the start time is outside the work hours, modify the start time to next nearest start time.

2nd function:

FUNCTION modify_finish_time (p_in_dte DATE) RETURN DATE
----------------------------------
IF p_in_dte - TRUNC (p_in_dte) > (19 / 24)
THEN
   RETURN TRUNC (p_in_dte) + (19 / 24);
ELSIF p_in_dte - TRUNC (p_in_dte) < (10 / 24)
THEN
   RETURN TRUNC (p_in_dte) - 1 + (19 / 24);
ELSE
   RETURN p_in_dte;
END IF;

If the finish time is outside the work hours, modify it to the previous nearest finish time.

3rd function:

FUNCTION get_days_to_exclude (p_in_start_date     DATE,
                              p_in_finish_date    DATE) RETURN NUMBER
--------------------------------------------------------
WITH cte --get all days between start and finish date
     AS (    SELECT p_in_start_date + LEVEL - 1 dte
               FROM DUAL
         CONNECT BY LEVEL <= p_in_finish_date + 1 - p_in_starT_date)
SELECT COUNT (1) * 9 / 24    --mutiply the days with work hours in a day
  INTO l_num_holidays
  FROM cte
 WHERE    TO_CHAR (dte, 'dy') = 'sun'    --find the count of sundays
       OR dte IN     --fins the count of holidays, assuming leaves are stored in separate table
             (SELECT leave_date  
                FROM leaves
               WHERE leave_date BETWEEN p_in_start_date
                                    AND p_in_finish_date);

l_num_holidays :=
   l_num_holidays + ( (p_in_finish_date - p_in_start_date) * (15 / 24)); --also, if the dates span more than a day find the non working hours.

RETURN l_num_holidays;

This function finds the no of days to be excluded while calculating the duration.

So, the final query should be something like this,

SELECT start_time,
       finish_time,
       CASE
          WHEN work_duration < 0 THEN NUMTODSINTERVAL (0, 'day')
          ELSE NUMTODSINTERVAL (work_duration, 'day')
       END
  FROM (SELECT start_time, finish_time,
               --modify_start_time (start_time), modify_finish_time (finish_time),
                 modify_finish_time (finish_time)
               - modify_start_time (start_time)
               - get_days_to_exclude (
                    TRUNC (modify_start_time (start_time)),
                    TRUNC (modify_finish_time (finish_time)))
                  work_duration
          FROM timings);

If the duration is less than 0, ignore it by setting it to 0.

Upvotes: 2

haki
haki

Reputation: 9779

You can take the hour difference between the two dates and reduce the amount of non-working hours times the days between the two days. consider this example : 

create table t1 (
    start_hour date , finish_hour date);

insert into t1 values (
  to_date('01/01/2013 12:00:00','dd/mm/yyyy hh24:mi:ss'),
  to_date('01/01/2013 15:00:00','dd/mm/yyyy hh24:mi:ss'));

insert into t1 values (
  to_date('01/01/2013 12:00:00','dd/mm/yyyy hh24:mi:ss'),
  to_date('02/01/2013 15:00:00','dd/mm/yyyy hh24:mi:ss'));

insert into t1 values (
  to_date('01/01/2013 18:00:00','dd/mm/yyyy hh24:mi:ss'),
  to_date('03/01/2013 11:00:00','dd/mm/yyyy hh24:mi:ss'));


with x as (
select start_hour ,finish_hour , 
      (finish_hour - start_hour) * 24 hour_diff , 
       trunc(finish_hour) - trunc(start_hour) as day_diff
from t1
) 
select  start_hour , finish_hour , 
        hour_diff - 15 * day_diff as working_hours 
from x;

|                     START_HOUR |                    FINISH_HOUR | WORKING_HOURS |
-----------------------------------------------------------------------------------
| January, 01 2013 12:00:00+0000 | January, 01 2013 15:00:00+0000 |             3 |
| January, 01 2013 12:00:00+0000 | January, 02 2013 15:00:00+0000 |            12 |
| January, 01 2013 18:00:00+0000 | January, 03 2013 11:00:00+0000 |            11 |

Upvotes: 0

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