On Fri, 2007-02-02 at 19:02 -0600, Richard Lynch wrote:
> On Fri, February 2, 2007 9:11 am, Dan Shirah wrote:
> > I am trying to populate a dropdown list to contain the current year to
> > 10
> > years in the future. Below is the code I'm using:
> >
> > <?PHP
> > for ($y=0;$y<=10;$y++) {
> > $years=date <http://php.net/date>('Y')+$y;
> > $short_years=date <http://php.net/date>('y')+$y;
> > echo "$short_years";
> > echo "<option value=\"$short_years\">$years</option>";
> > }
> > ?>
> > I want the selection value to be 2007, 2008, 2009, 2010.... and the
> > $years
> > accomplishes this. Howeverm I want the option value to be the two
> > digit
> > year ex. 07, 08, 09, 10...the $short_years semi accomplishes
> > this....instead
> > of getting 07, 08, 08, 10 I get 7, 8, 9, 10...how can I get it to
> > output the
> > two year for 07, 08, 09 without it cutting off the zero?
>
> for ($year = date('Y'); $year <= date('Y') + 10; $year++){
> $short_year = $year - 2000;
> echo "<option value=\"$short_year\">$year</option>\n";
> }
>
> Using the 2-digit year as your value is almost for sure a really Bad
> Idea, unless you are dealing with legacy software that absolutely
> needs it...
Sorry Richard, you solution fails since he wants to retain the leading
zeros on the 2 digit years. Using subtraction and not formatting the
result will result in the loss of the 0 in 07. Also, why call the date()
function 10 times by embedding it in the loop's terminate condition?
Cheers,
Rob.
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