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
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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:
> >
> >
> > > chdir("/home/casapu/paginas/images/carteleras");
> > $direc = opendir(".");
> > while ($f = readdir($direc)); {
> > print("".$f."");
> >}
> > ?>
> >.");
> >
> > 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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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:
> >
> >
> > > chdir("/home/casapu/paginas/images/carteleras");
> > $direc = opendir(".");
> > while ($f = readdir($direc)); {
> > print("".$f."");
> >}
> > ?>
> >.");
> >
> > 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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To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP] Trying to list a directory content HELP PLEASE
This is not how readdir() works, it takes on
a directory resource from opendir()
http://www.php.net/readdir
I'm guessing you meant to put the directory
name into opendir() which will work, I don't
see any advantage to using chdir() here.
Maybe it's permissions or doesn't exist,
consider using a few other functions in there
such as is_dir() and file_exists().
Regards,
Philip Olson
On Thu, 6 Jun 2002, Scott Hurring wrote:
> 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:
> > >
> > >
> > > > > chdir("/home/casapu/paginas/images/carteleras");
> > > $direc = opendir(".");
> > > while ($f = readdir($direc)); {
> > > print("".$f."");
> > >}
> > > ?>
> > >.");
> > >
> > > 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
>
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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:
> >
> >
> > > chdir("/home/casapu/paginas/images/carteleras");
> > $direc = opendir(".");
> > while ($f = readdir($direc)); {
> > print("".$f."");
> >}
> > ?>
> >.");
> >
> > 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
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:
>
>
>chdir("/home/casapu/paginas/images/carteleras");
> $direc = opendir(".");
> while ($f = readdir($direc)); {
> print("".$f."");
>}
> ?>
>.");
>
> 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] 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: ".$f.""); } ?> ."); 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
