Calculate the hours between two dates

Hello

I have a requirement to count the hours with 2 given timestamps, something like:
SQL> select ((to_date('22-oct-2012 13:00:00','dd-mon-yyyy hh24:mi:ss') - to_date('20-oct-2012 13:00:00','dd-mon-yyyy hh24:mi:ss'))*24) B from dual ;

         B
----------
        48
However, I wish it would be that simple. I just need to count the hours between two timestamps for the hours of work for example, 08:00 - 17:00, MONDAY to Friday (excluding weekends).
So lets say I have to count the hours between two dates above the result should be 18 hours not 48 (4 hours on 20/oct, 9 hours on 21/oct to 5 hours 22/oct).
How can I achieve this? Any idea will be useful.

Thank you

My query generates a list of all the hours between the start and end time, filters on the hours that are outside of the hours of work and then account for the remaining lines. Change that to partial hours must just be a case of replace the two occurrences of 1/24 with 24/01/60, then - as say you - Division COUNT (*) by 60.

There may be performance issues using techniques of generation of line like this, if you put this in a query involving the joining of several lines of other tables (for example where the dates of beginning and end are derived from several employees?) or in several days. If you don't find it then count the number of whole days (taking weekends into account) between dates and work on the partial days after that it could be better. This would probably involve a lot of statements of case (which I don't have time to test it now).

Ben

Tags: Database

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  • Calculate business hours between two dates in a cursor

    Hello, its me again (?).

    Yesterday, I got a solution to a problem here where I need to calculate the hours between two dates, excluding Sat/Sun and from 09:00 to 18:00. I'm now putting this login in a PL/SQL block, so I can calculate it for each record in a slider, all by looping.
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       ,end_date
       )
as (select  to_date('22-oct-2012 09:00:00','dd-mon-yyyy hh24:mi:ss') start_dt,
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SELECT  to_date(c_record.INICIOREAL,'dd-mon-yyyy hh24:mi:ss') INTO vFch2 FROM DUAL;

IF ((vHoras > 2) AND c_record.PRIORIDAD=3) THEN
  vContOk :=vContOk + 0;
  vContBad := vContBad+1;
  vContOt   := vContOt+1;
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  vContOk :=vContOk +1;
  vContBad := vContBad+0;
  vContOt   := vContOt+1;
END
    IF;

IF ((vHoras > 4) AND c_record.PRIORIDAD=2) THEN
  vContOk :=vContOk + 0;
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  vContBad := vContBad+0;
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END IF;

IF ((vHoras > 16) AND c_record.PRIORIDAD=1) THEN
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END LOOP;
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  4                                ) work_minutes
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  6  /

WORK_MINUTES
------------
         261

SQL>

    SY.

  • Excerpt from the hours between two dates

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    Post edited by: Parth272025

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        END;
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    Concerning
    Patrick
    -----------
    My Blog: http://www.inside-oracle-apex.com
    APEX 4.0 Plug-Ins: http://apex.oracle.com/plugins
    Twitter: http://www.twitter.com/patrickwolf

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    order_type

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    overtime

    of glas_order o

    model

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    main m

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    then presentv (r.b [wday [0]],)

    less (order_ready [0], to_date (to_char (order_ready [0], 'YYYYMMDD') | r.e [WDAY [0]],'YYYYMMDDHH24:mi'))))

    -to_date (to_char (order_ready [0], 'YYYYMMDD') | r.b [WDAY [0]], 'YYYYMMDDHH24:MI')

    0) * 24

    of another nvl (r.dur [wday [0]], 0)

    end

    )

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    2 22 Order1 17
    3 10 Order2 2

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    , comics [0] = to_date (to_char (order_date [0] + iteration_number, 'YYYYMMDD') | r.b [WDAY [0]], 'YYYYMMDDHH24:MI')

    , ed [0] = to_date (to_char (order_date [0] + iteration_number, 'YYYYMMDD') | r.e [WDAY [0]], 'YYYYMMDDHH24:MI')

    , d [0] = d [0] +.

    case

    When trunc(order_date[0]) = trunc(order_ready[0]) - beginning and ready same day

    then presentv (r.b [wday [0]],)

    less (largest (order_ready [0], [0] bd), ed [0])

    -Greatest (BD [0], least(ORDER_DATE[0],ED[0]))

    0) * 24

    When order_date [0] + iteration_number = order_date [0] - first day

    then presentv (r.b [wday [0]],)

    ED [0]

    -Greatest(ORDER_DATE[0],BD[0])

    0) * 24

    When trunc(order_date[0]) + iteration_number = trunc(order_ready[0]) - last day

    then presentv (r.b [wday [0]],)

    least(order_ready[0],ED[0])

    -bd [0]

    0) * 24

    of another nvl (r.dur [wday [0]], 0)

    end

    )

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