$x = fopen ressource as as get parameter.

[Bob Bedford]

Hi all,

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

$x=fopen($filen ame);
header('locatio n:myfile.php?re ssource='.$x);

but it doesn't seem to work.

How to do so ?

Bob

[Erwin Moller]

Bob Bedford wrote:

Hi all,

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

$x=fopen($filen ame);
header('locatio n:myfile.php?re ssource='.$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

[chonny]

fopen is function that open files.
see this: http://php.net/manual/en/function.fopen.php
Bob Bedford напи�а:

Hi all,

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

$x=fopen($filen ame);
header('locatio n:myfile.php?re ssource='.$x);

but it doesn't seem to work.

How to do so ?

Bob

[Pedro Graca]

Bob Bedford wrote:

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

$x=fopen($filen ame);
header('locatio n:myfile.php?re ssource='.$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('Locatio n: http://www.yourserver. com/path/to/myfile.php?file name=' . urlencode($file name));
exit(0);
myfile.php
// validate $_GET['filename']
$f = fopen($_GET['filename']);

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

[Bob Bedford]

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

Bob