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?

@BartFriederichs for some reason, I'm sure, else I wouldn't be asking... That's not important. The point is I am asking why this won't work this way. I don't want to run it by a different code structure. — Fane
– Fane, Commented Aug 14, 2015 at 12:57

Bart Friederichs · Accepted Answer · 2015-08-14 11:36:05Z

2

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.

answered Aug 14, 2015 at 11:36

Bart Friederichs

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3 Comments

Fane Over a year ago

I echo the output. Nothing is shown. This is called as soon as my server starts. No problem in performance.

Fane Over a year ago

none. I only get an error when I change the path and get "could not open/find file xxx/another_script.php". Since I get nothing like this I'm assuming it is correct...

Fane Over a year ago

so, no idea? I really can't figure it out at all, do you think of anything that might be off? :)

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