Can't run shell_exec(.php) with mysql in php script

Question

Different from the other situations I found here on SO

I have the following code in the main php script

include("sql.php"); //holds all the data to mysql (and variable $db to connect)

$output = shell_exec('/usr/local/bin/php folder/another_script.php')

and in another_script.php I have this:

include_once("../sql.php"); //notice difference to previous include in the other script

$query = "some query";
$db->query($query)

$output in the first script is blank. The second script runs when alone and not called from another script. What am I missing?

Answer 1

The documentation says:

Execute command via shell and return the complete output as a string

It returns the output. You never use the output, hence nothing is shown. You could do this:

$output = shell_exec('/usr/local/bin/php folder/another_script.php')
echo $output;

Yet, I advise you to reconsider your design to create classes or functions to separate functionality instead of calling (quite expensive) shells to call PHP code.