Chris,
A pointer is a type of shark! Just like them, this suggestion has sharp
teeth too - change the time_in value to 11:45 and see what I mean...
This topic was discussed one week ago ("Time Calculations") - see archives
(to save you that, I'll forward you a copy privately (list doesn't like
attachments)).
Regards,
=dn
> Thanks for the quick response. The times are stored in TIME format. If I
> do the following:
>
> SELECT *, time_out-time_in as job_time from time_sheet;
>
> the job_time column gives me the right time, but is not formatted, for
> example:
>
> time_in = 11:00
> time_out = 13:15
>
> job_time= 21500
>
> Any pointers?
>
> -Chris
>
> On Saturday, August 10, 2002, at 09:36 AM, DL Neil wrote:
>
> > Hello Christopher,
> >
> >> I need to ask for some help and I can't find anything in the manuals. I
> >> am writing a php based app that as one of it's functions keeps an
> >> employee time sheet. The times are stored in a mysql database with 2
> >> columns - time_in and time_out. What I need to do is subtract these to
> >> columns to show a job_time (which should be in hh:mm form).
> >>
> >> Can anyone give me any suggestions? Or am I better off doing this on
> >> the
> >> PHP side rather than the mysql side?
> >
> >
> > My general rule is that when pulling data out of an RDBMS it is best to
> > do
> > as much as possible there, than to transfer data into (say) PHP and
> > post-process it.
> >
> > Are the times stored as DATETIME/TIME, timestamp, or other format data?
> > The
> > answer will thus vary! Assuming we're not talking about timestamps
> > (trivial
> > exercise) then SEC_TO_TIME() and its counterpart may be of interest.
> > Manual
> > ref: http://www.mysql.com/doc/en/Date_and_time_functions.html
> >
> > Regards,
> > =dn
> >
> >
> >
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