$x = fopen ressource as as get parameter.

Hi all,

I'm trying to pass a fopen() ressource result as a parameter.

$x=fopen($filename);
header('location:myfile.php?ressource='.$x);

but it doesn't seem to work.

How to do so ?

Bob

Bob Bedford wrote:

> Hi all,
>
> I'm trying to pass a fopen() ressource result as a parameter.
>
> $x=fopen($filename);
> header('location:myfile.php?ressource='.$x);
>
> but it doesn't seem to work.
>
> How to do so ?
>
> Bob


Hi Bob, you don't.
A resource is something you can create for the duration of your script, so
when your script finishes, it is gone.

In your case I would pass the filename, and let the receiving script make
its own resource based on the filename.

Regards,
Erwin Moller

fopen is function that open files.
see this: http://php.net/manual/en/function.fopen.php


Bob Bedford написа:
> Hi all,
>
> I'm trying to pass a fopen() ressource result as a parameter.
>
> $x=fopen($filename);
> header('location:myfile.php?ressource='.$x);
>
> but it doesn't seem to work.
>
> How to do so ?
>
> Bob

Bob Bedford wrote:
> I'm trying to pass a fopen() ressource result as a parameter.
>
> $x=fopen($filename);
> header('location:myfile.php?ressource='.$x);
>
> but it doesn't seem to work.

When the script ends (after the header() call) the file gets closed and
the resource terminated. There is no way to pass a resource (any
resource, not just a open file handle) to other scripts; you can't do it
with URL parameters nor with POSTed data nor with cookies nor with
session variables.

> How to do so ?

Pass the name of the file and reopen it.

header('Location: http://www.yourserver.com/path/to/myfile.php?filename=' . urlencode($filename));
exit(0);


myfile.php
// validate $_GET['filename']
$f = fopen($_GET['filename']);

--
File not found: (R)esume, (R)etry, (R)erun, (R)eturn, (R)eboot

Thanks for your help....didn't find out so I guessed it wasn't possible

Bob