$argv is an array. Don't expect if you coded in C that this is pointer to the first so 
you possibly want to do echo of the first in
the array.
instead do :
foreach($argv as $param_name => $param_value){
// do your stuff here

Andrey Hristov
IcyGEN Corporation

----- Original Message -----
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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