Original Post
Hi, I'd like to process another php file from a php script. My first solution goes like this:
$cmd = "C:\\...\\apachefriends\\xampp\\php\\php.exe -f test.php";
exec($cmd, $out);
for ($i=0; $i<count($out); $i++)
echo $out[$i];
But I have some problems with this solution: I need to know where the php.exe is located on the server. Is there some kind of global variable containing this path? When the exec() command is executed a terminal popps open and closes again. Because I have a apache-server on my system I'd like to know if the window will be seen at the server side or at the client side? Is this a good/elegant solution at all? because the exec() command will be called every time a user navigates on my site. So every time a window will popp open on the server/client side! Is ther a better way to do this?? Thanks in advance!! Constantin