RE: [PHP] Trying to list a directory content HELP PLEASE

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 ScottH@eac.com Voice: 201-462-2149 Fax: 201-288-1515 > -----Original Message----- > From: Jason Wong [mailto:php-general@gremlins.com.hk] > Sent: Thursday, June 06, 2002 3:44 PM > To: php-general@lists.php.net > 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 >