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 as you told me...
$fd=readdir(/home/casapu/paginas /image/caterleras/);
if (!$fd) die (Can't read dir);
It gives me:
Warning: Supplied argument is not a valid Directory resource in
/home/casapu/paginas/mbt/php/dir.php on line 7
Can't read dir
So I'm assuming is not accepting the directory. I have checked the
permits, and it has all enabled, read, write and executable.
I have tried with the final slash, and without it, and so far it keeps
giving me that message... any ideas?
-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 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
*/
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RE: [PHP] Trying to list a directory content HELP PLEASE
Is that a space in the directory path?
I think you will have to use opendir() function first.
opendir: Returns a directory handle to be used in subsequent closedir(),
readdir(), and rewinddir() calls.
The manual has some good examples.
-Original Message-
From: [EMAIL PROTECTED]
[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 as you told me...
$fd=readdir(/home/casapu/paginas /image/caterleras/);
if (!$fd) die (Can't read dir);
It gives me:
Warning: Supplied argument is not a valid Directory resource in
/home/casapu/paginas/mbt/php/dir.php on line 7
Can't read dir
So I'm assuming is not accepting the directory. I have checked the
permits, and it has all enabled, read, write and executable.
I have tried with the final slash, and without it, and so far it keeps
giving me that message... any ideas?
-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 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
*/
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP] Trying to list a directory content HELP PLEASE
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: Supplied argument is not a valid Directory resource in /home/casapu/paginas/mbt/php/dir.php on line 7 Can't read dir So I'm assuming is not accepting the directory. I have checked the permits, and it has all enabled, read, write and executable. I have tried with the final slash, and without it, and so far it keeps giving me that message... any ideas? Does the directory paginas really have a blank space at the end of the name? miguel -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Trying to list a directory content HELP PLEASE
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?
HELP PLEASE
VV
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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
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
[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
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