exec script and how to pass a parameter/varE

I Want to execute another PHP script and want to pass a parameter/variable.
Tried to pass as a get this way:
$var="test";
$execute="/home/myaccount/special/sda51.cgi?var=$var";
exec("/usr/bin/php $execute");

It does not execute so I tried the following which does execute but I cannot grab the var

$var="test";
$execute="/home/myaccount/special/sda51.cgi";
exec("/usr/bin/php $execute $var");

Tried to the following without any results:
$var1=$var['0'];
$var1a=$var['1'];
$var1b=$var['2'];
$var1c=$var['3'];
$var1d=$var['4'];
$var2=$var['var'];


Any Ideas ?
Thanks,
Peter M

is there a specific reason why you can't just include() it?

$execute="/home/myaccount/special/sda51.cgi?var=".$var; ?

When I use include(
The included script will not exec( the next script.

As soon as I place a ? in the following, there is no execution at all.

$execute="/home/myaccount/special/sda51.cgi?var=".$var; ?

Can you think of any other tricks ?

Many thanks,
Peter

Could you maybe explain to us what the scenario is (bigger picture), maybe there is a completely different (possibly more efficient) way of skinning it?

Instead of executing the php file directly, write a batch file with all the command line parameters...

First, drop to your command line interface and try to execute the script using php as a command line call... IF that works, then take that exact line and have your program write a batch file (i.e. myfile.bat) and then call that to execute... then delete it...

Although, you should be able to directly invoke a command line execution of PHP from within a PHP script.

So, I would first run it manually from the command line to ensure it is working...

Cheers,
Keith.

You are aware that the ? should not be included in .$var; ?

It should just be .$var;
The ? was because I wasn't sure it might solve the problem

u can use argc and argv to access command line parameters, but i think they are now in $_SERVER array.

anyway, look at http://php.net/features.commandline for more...