Thanks Justin, your solution is spot-on!
Regards
Steve
Justin French wrote:
> <SELECT name="day">
> <OPTION value="please select" <? if(empty($day)) { echo "selected"; }
> ?>>
> <OPTION value="monday" <? if($day == "monday") { echo "selected"; } ?>>
> <OPTION value="tuesday" <? if($day == "tuesday") { echo "selected"; }
> ?>>
> <OPTION value="wednesday" <? if($day == "wednesday") { echo "selected";
> } ?>>
> <OPTION value="thursday" <? if($day == "thursday") { echo "selected"; }
> ?>>
> <OPTION value="friday" <? if($day == "friday") { echo "selected"; } ?>>
> </SELECT>
>
> Obviously this is labourios to code... you can do this a lot smarter/quicker
> with an array for the entire select box... have an array of days, and do a
> foreach loop which writes the all the options for you, with the if
> statements, etc etc.
>
> By the way, this has nothing to do with sessions :)
>
> It's purely about how to populate drop-down menus from an array, and how to
> have the correct value selected if it exists, else showing a default
> selection.
>
> Cheers,
>
> Justin French
>
> on 08/07/02 9:36 PM, Steve Fitzgerald ([EMAIL PROTECTED]) wrote:
>
> > I am designing a form using sessions in which the user inputs their
> > details on page 1 and after submitting they are directed to page 2 for
> > confirmation. They then have the option of editing their input (ie they
> > are returned to page 1) where their previous input is reflected in the
> > form fields by
> > value ='<?php echo $var ?>'
> > This works fine except if the input type is a drop down box, in which
> > case the default <option> is shown. Is there any way around this? How
> > can I show the user their previous choice in these boxes?
> > Any insights would be appreciated.
> > Steve
> >
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