Ok, I am using this modified code from the
manual. It works great, but how do I modify
it further to only show .jpg files ?
<snip>
<?php
$handle=opendir('.');
echo "<select name=image_url>";
while (false!==($file = readdir($handle))) {
if ($file != "." && $file != "..") {
echo "<option ";
echo "value=".$file.">";
echo $file;
echo "</option>\n\r";
}
}
echo "</select>";
closedir($handle);
?>
</snip
Jerry Lake - [EMAIL PROTECTED]
Web Designer
Europa Communications - http://www.europa.com
Pacifier Online - http://www.pacifier.com
-----Original Message-----
From: Sean R. Bright [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 07, 2001 4:50 PM
To: 'Jerry Lake'; 'Henrik Hansen'; 'php general'
Subject: RE: [PHP] populate select box with contents of a file?
Take a look at dir() and readdir(). Those along with echo() should help you
out.
Sean
> -----Original Message-----
> From: Jerry Lake [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, March 07, 2001 7:40 PM
> To: Henrik Hansen; php general
> Subject: [PHP] populate select box with contents of a file?
>
>
> I would like to be able to populate
> the options of a select box with the
> contents of my images directory online
> so I can select the image I want to
> go with the form I am filling out.
> what functions do I need to look into
> to figure this one out?
>
> Jerry Lake - [EMAIL PROTECTED]
> Web Designer
> Europa Communications - http://www.europa.com
> Pacifier Online - http://www.pacifier.com
>
>
>
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