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Vieux 28/08/2007, 03h18   #3
Peter Brawley
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Par défaut Re: [MYSQL]Time formatting for cycle time.

Craig,

>I am working on Martin Minka's date diff function as found
>at http://forge.mysql.com/snippets/view.php?id=56. It is a beautiful
>thing. However, I am trying to alter it or identify a similar function
>that instead of giving me the number of days between two dates it
>returns the number of workday hours:minutes between two datetimes, or
>some other date differential (such as an exact number of days between
>two dates with remainder)


Here's a logically equivalent datediff calc, mebbe slightly simpler:

SET @d1 = '2007-1-1';
SET @d2 = '2007-3-31';
SET @dow1 = DAYOFWEEK(@d1);
SET @dow2 = DAYOFWEEK(@d2);
SET @days = DATEDIFF(@d2,@d1);
SET @wknddays = 2 * FLOOR( @days / 7 ) +
IF( @dow1 = 1 AND @dow2 > 1,
1,
IF( @dow1 = 7 AND @dow2 = 1, 1,
IF( @dow1 > 1 AND @dow1 > @dow2, 2,
if( @dow1 < 7 AND @dow2 = 7, 1, 0 )
)
)
);
SELECT FLOOR(@days-@wkndDays) AS BizDays;

To include time in the difference, you could adopt as a return
convention a string format like 'N days hh:mm:ss', where N is the date
difference calculated above, minus one if the time portion of d1 is
later than than that of d2. Something like this:

SET @d1 = '2007-1-1 00:00:00';
SET @d2 = '2007-3-31 12:00:00';
SET @tdiff = TIMEDIFF( TIME(@d1), TIME(@d2) );
SET @dow1 = DAYOFWEEK(@d1);
SET @dow2 = DAYOFWEEK(@d2);
SET @days = DATEDIFF(@d2,@d1);
SET @wknddays = 2 * FLOOR( @days / 7 ) +
IF( @dow1 = 1 AND @dow2 > 1,
1,
IF( @dow1 = 7 AND @dow2 = 1, 1,
IF( @dow1 > 1 AND @dow1 > @dow2, 2,
IF( @dow1 < 7 AND @dow2 = 7, 1, 0 )
)
)
);
SET @days = FLOOR(@days - @wkndDays) - IF( @tdiff < 0, 1, 0 );
SET @tdiff = IF( ASCII(@tdiff) = 45, SUBSTRING(@tdiff,2), TIMEDIFF(
'24:00:00', @tdiff ));
SELECT CONCAT( @days, ' days ', @tdiff );

PB

-----

Weston, Craig (OFT) wrote:
> Hello all,
>
> I am working on Martin Minka's date diff function as found
> at http://forge.mysql.com/snippets/view.php?id=56. It is a beautiful
> thing. However, I am trying to alter it or identify a similar function
> that instead of giving me the number of days between two dates it
> returns the number of workday hours:minutes between two datetimes, or
> some other date differential (such as an exact number of days between
> two dates with remainder)
>
>
>
>
>
> I hate asking open ended questions, but can anyone give me any hints as
> to how to make this conversion? I may be able to figure it out... In a
> few weeks.
>
>
>
> My long term goal is to identify the business hours(minutes, seconds,
> whatever) between two dates, taking into account weekends, holidays, and
> business hours.
>
>
>
> Thanks,
>
> Craig
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