?
-Mensaje original-
De: Scott Hurring [mailto:[EMAIL PROTECTED]]
Enviado el: Jueves, 06 de Junio de 2002 15:26
Para: '[EMAIL PROTECTED]'
Asunto: 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
]
[mailto:[EMAIL PROTECTED]
t]On Behalf Of webmaster mbtradingco
Sent: Thursday, June 06, 2002 3:56 PM
To: 'Scott Hurring'
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] Trying to list a directory content HELP PLEASE
Hey Scott, that at least helped me to find out what is going wrong.
When I use the code
On Thu, 6 Jun 2002, webmaster mbtradingco wrote:
Hey Scott, that at least helped me to find out what is going wrong.
When I use the code as you told me...
$fd=readdir(/home/casapu/paginas /image/caterleras/);
if (!$fd) die (Can't read dir);
It gives me:
Warning:
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
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
]
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
6 matches
Mail list logo
