var_dump($argv);
$argv is an array. Don't expect if you coded in C that this is pointer to the first so
you possibly want to do echo of the first in
the array.
instead do :
foreach($argv as $param_name => $param_value){
// do your stuff here
}
Regards,
Andrey Hristov
IcyGEN Corporation
http://www.icygen.com
BALANCED SOLUTIONS
----- Original Message -----
From: "PACKER, Steffan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, November 23, 2001 7:02 PM
Subject: [PHP] argv
> the $argv is meant to contain all the arguments sent to the script. I need
> to get access to these but when I try to access them I can't!
> if I just use echo($argv) the only output is 'Array'
>
> All I am expecting to find for the purpose of my script is a five letter
> code, not a name/value pair.
>
> Can anyone help?
>
> Thanks,
> Steffan
>
>
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