如何将参数传递给脚本

Question

I am "forking" php code, kicking off scripts to run in the background allowing the browser application to continue without having to wait for a response.

I would like to be able to pass a parameter to such a script.

Say i have an array i would like to be accessible to the script in question, how could i pass it over?

$test = array(1,2,3);

//for windows i do the following which works well enough 
pclose(popen("start php /path/to/script.php ","r"));

OR for some linux distro's :

exec('php -f /path/to/script.php >/dev/null 2>&1 &');

so firstly, is it possible to pass my array to that script? if its possible to make it accessible to the script, how could i access the passed variable in that script?

(sitting here confused, thinking about passing variable by value/reference....)

thanks!

Answer 1

Nicolas Brown is almost right. But that would be the second part of the solution you're looking for, I think.
As you maybe already know you can call a program from the console and yet pass any arguments the program might expect:

>php -f /path/to/program arg1[ arg2 ... argN] >/dev/null 2>&1 &

So as you're passing a string to exec() why not just "insert" your array as arguments and then execute it

$test = array(1,2,3);
$args = implode($test, " ");
$prog = 'php -f /path/to/script.php ' . $args . ' >/dev/null 2>&1 &'
exec($prog);

Here comes something from the tutorial Nicolas Brown mentioned. In your script.php you can get the parameters from $argv[]

$myParam1 = $argv[1];
$myParam2 = $argv[2];

Please, note that in argv[0] is the name of the program. That's exactly why in the tutorial they first do arry_shift($argv), to get rid off the first element (the program's name and path).
Hope it helps!