使用参数从另一个调用PHP脚本并获取输出

Question

I have a php script, let's call it one.php. This calls a second php script two.php with some arguments. Right now my code in one.php looks like this:

shell_exec("php ./two.php" . ' ' . escapeshellarg($var1) . ' '.      
escapeshellarg($var2) . ' '. escapeshellarg($var3));

This works fine (although I don't like it), but now I want to get some values that are computed in two.php.

I colud use include() or require() to get access to all the variables, but I want to use two.php in more than just one script - without having to use the same variable names in every script that calls two.php (which I had to because of the scope of the variables in include()).

My php skills are quite poor, so: Is there a way to send variables of a called script to the calling script or something like this? Or maybe another way of calling two.php.

Update: Using a function did it. Thanks!

Answer 1

"Calling another script" via shell_exec is a pretty bad idea. Instead, two.php should contain a bunch of functions or classes which you can include and call anytime:

include 'two.php';

$data = func_from_two($var1, $var2, $var3);

Learn to program using functions and classes, that's the only sane way to go.