On May 9, 4:52*pm, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> Mo wrote:
> > On May 9, 10:03 am, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> >> Mo wrote:
> >>> On May 8, 5:03 am, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> >>>> Álvaro G. Vicario wrote:
> >>>>> Mo escribió:
> >>>>>> On the time report, we want to see EVERY date under each employee, and
> >>>>>> the string "Absent" for the dates on which the user has no punch
> >>>>>> times. (It would also be nice to include day-names on the report.)
> >>>>>> Obviously the Time table will only have entries for when people
> >>>>>> actually clock in/out, so how do I get the other dates to show on the
> >>>>>> report?
> >>>>>> Any suggestions?
> >>>>>> One guy here suggested just adding a table (to the MySQL DB ) which
> >>>>>> includes every date from here to ...., then just do an outer join.
> >>>>> That's crazy. You'd be overpopulating your DB with dumb data that will
> >>>>> need forever maintenance.
> >>>>> It's way easier than that:
> >>>>> 1. Fetch the available DB data into a PHP array
> >>>>> 2. In PHP, loop through the range of dates. For each date:
> >>>>> * * - If there is data, print it
> >>>>> * * - If not, print 'Absent'
> >>>>> SELECT UNIX_TIMESTAMP(DATE) AS DATE, USER, BLAH
> >>>>> FROM PUNCH_TIMES
> >>>>> WHERE .....
> >>>>> As about looping:
> >>>>> $from = mktime(0, 0, 0, 1, 1, 2008);
> >>>>> $to = time();
> >>>>> for($i=$from; $<=$to; $i=strtotime('+1 day', $i)){
> >>>>> * * echo date('d/m/Y', $i) . ":\n";
> >>>>> * * if( isset($punch_data[$employee][$i]) ){
> >>>>> * * * * ...
> >>>>> * * }else{
> >>>>> * * * * echo "Absent\n";
> >>>>> * * }
> >>>>> * * /*...*/
> >>>>> }
> >>>>> I'm missing several steps but I hope you see the idea.
> >>>> No, it's not crazy, and it's quite commonly done - I've seen it in a
> >>>> bunch of different RDBMS's.
> >>>> Doing it like this and everything can be handled in one SQL SELECT
> >>>> statement. *Not only will it be faster, but there's much less code to
> >>>> maintain.
> >>>> --
> >>>> ==================
> >>>> Remove the "x" from my email address
> >>>> Jerry Stuckle
> >>>> JDS Computer Training Corp.
> >>>> jstuck...@attglobal.net
> >>>> ==================- Hide quoted text -
> >>>> - Show quoted text -
> >>> Of course I always preffer speed of execution and simplicity of code,
> >>> but I've just come up with one concerne ...
> >>> Would we be able to generate a report (without errors) across a
> >>> NewYear date if we were numbering the days 1-366?
> >>> We have one group of our employees which gets paid on every 15th, so
> >>> even in our regular reports, we would encounter this at least once per
> >>> year.
> >>> ~Mo
> >> No, you just need to check to see if this is a leap year or not.
>
> >> --
> >> ==================
> >> Remove the "x" from my email address
> >> Jerry Stuckle
> >> JDS Computer Training Corp.
> >> jstuck...@attglobal.net
> >> ==================- Hide quoted text -
>
> >> - Show quoted text -
>
> > So, if I generate a report this year (2008) for the 12/16/07 - 1/15/08
> > date range, how do I make it identify the day numbers from December as
> > being from a prior year?
> > If I'm understanding all this correctly, our range of day numbers
> > would be 349-15.
> > Would I just have to use some function to extract the year value from
> > the date range provided by the user via the form?
>
> > ~Mo
>
> At this point you're getting farther out of the PHP range and more into
> MySQL. *You need to be asking this in comp.databases.mysql.
>
> It can be done - but this isn't a PHP problem.
>
> --
> ==================
> Remove the "x" from my email address
> Jerry Stuckle
> JDS Computer Training Corp.
> jstuck...@attglobal.net
> ==================- Hide quoted text -
>
> - Show quoted text -

cool, thanks