php parameter usage in Unix (beginner question)

Consider this php script test.script, and the call: php test.script 4444

<?php
$card = $argv[ 1 ];
echo "$card\n";
require_once 'HTTP/Request.php';
$req = new HTTP_Request( 'http://www.???.com?cc_number=4444&exp_date=0612' );
$req -> sendRequest();
?>

(Note that I've omitted the actual URL I'm testing with, as that isn't a factor).

The echo command correctly displays 4444 to my screen, but my goal here is to use the passed parameter in the HTTP_Reqest line, as in:

$req = new HTTP_Request( 'http://www.???.com?cc_number=$card&exp_date=0612' );

But using the "$card" syntax exactly how I have it in the line above does not work, when the hardcoded "4444" above does work properly. There must be something I do not yet understand about how to use variables in php, but I cannot find syntax examples anywhere. Can anyone direct me as to how I can utilize an input parameter as a variable in the HTTP call above? Any help is much appreciated.

 

Replace single quotes with double quotes

 

Its your use of quotes; if you use doouble quotes variables will be expanded, in single they will not.

consider this.

<?php
$card = 4444;

print 'hello $card';
print "hello $card";
?>

PHP:

OUTPUT:
hello $card
hello 4444

You can get away with it by concatenating the value;

$req = new HTTP_Request( 'http://www.???.com?cc_number='.$card.'&exp_date=0612' );

PHP:

 

Thanks - your suggestions work. It's tough being a newbie