Question

I am wanting to call a php file using exec.

When I call it I want to be able to pass a variable through (an id).

I can call echo exec("php /var/www/unity/src/emailer.php"); fine, but the moment I add anything like echo exec("php /var/www/unity/src/emailer.php?id=123"); the exec call fails.

How can I do this?

Answer 1

Your call is failing because you’re using a web-style syntax (?parameter=value) with a command-line invokation. I understand what you’re thinking, but it simply doesn’t work.

You’ll want to use $argv instead. See the PHP manual.

To see this in action, write this one-liner to a file:

<?php print_r($argv); ?>

Then invoke it from the command-line with arguments:

php -f /path/to/the/file.php firstparam secondparam

You’ll see that $argv contains the name of the script itself as element zero, followed by whatever other parameters you passed in.