Instead of chdir() try putting the path directly into
readdir(); it'll make the code a tiny bit cleaner.
readdir("/home/casapu/paginas/images/carteleras");
** and check return values! **
$fd = readdir(...)
if (!$fd) die("Cannot readdir");
The code you have *should* work, but you'll never
know why it's not working if you don't check return
statuses
---
Scott Hurring
Systems Programmer
EAC Corporation
[EMAIL PROTECTED]
Voice: 201-462-2149
Fax: 201-288-1515
> -----Original Message-----
> From: Jason Wong [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, June 06, 2002 3:44 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] Trying to list a directory content HELP PLEASE
>
>
> On Friday 07 June 2002 00:47, webmaster mbtradingco wrote:
> > I know my doubt is probable odd, but I would ask your help please.
> >
> > I need a user to be able to select an image from a directory, from a
> > drop down box. For this I need to list all the images
> available on the
> > directory, hence, I have this code:
> >
> > <select size="1" name="normal1">
> > <?php
> > chdir("/home/casapu/paginas/images/carteleras");
> > $direc = opendir(".");
> > while ($f = readdir($direc)); {
> > print("<option value=\"".$f."\">".$f."</option>");
> > }
> > ?>
> > .</select>");
> >
> > but this is not working. I have reviewed the code against all the
> > manuals/books I have, and it says it should work but it
> doesn't. Anyone
> > knows what I'm doing wrong?
>
> how doesn't it work?
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
> Open Source Software Systems Integrators
> * Web Design & Hosting * Internet & Intranet Applications
> Development *
>
> /*
> Fashions have done more harm than revolutions.
> -- Victor Hugo
> */
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
