PHP exec command not having an $output but $return is 0

Question

This is my code for executing a command from PHP:

            $execQuery = sprintf("/usr/local/bin/binary -mode M \"%s\" %u %s -pathJson \"/home/ec2/fashion/jsonS/\" -pathJson2 \"/home/ec2/fashion/jsonS2/\"", $path, $pieces, $type);
            exec($execQuery, $output, $return);

the $return value is always 0 but $output is empty. The $output should be a JSON.

If I execute the same but removing one letter to binary (for example /usr/local/bin/binar ) I get (correctly) a $return = 127.
If I write other parameters (like -mode R which doesn't exit) I got errors from the console (which are correct as well).
If I run the exact $execQuery (which I printf before to be sure about quotation marks) on the console, it executes correctly. It's only the PHP side where I've got the error.

What can be wrong? Thank you in advance.

To clarify, $output is an empty array right?

Tivie
– Tivie

2015-02-12 17:04:50 +00:00
Commented Feb 12, 2015 at 17:04 — Tivie
– Tivie, Commented Feb 12, 2015 at 17:04
Yes, it's an empty array

Rafael Ruiz Muñoz
– Rafael Ruiz Muñoz

2015-02-12 17:11:36 +00:00
Commented Feb 12, 2015 at 17:11 — Rafael Ruiz Muñoz
– Rafael Ruiz Muñoz, Commented Feb 12, 2015 at 17:11

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

Well, a couple of things might be happening...

This binary you're running write to something else that STDOUT (for instance, STDERR)
The env vars available to the PHP user differ from the env vars available to the user running console (and those vars are required)
PHP User does not have permission to access some files involved.

In order to debug, it might be better to use proc_open instead of exec, and check the STDOUT and STDERR. This might give you additional information regarding what's happening.

Suggestion (and shameless advertising)

I wrote a small utility library for PHP that executes external programs in a safer way and provides aditional debug information. It might help you to, at least pinpoint the issue.

edited Jun 20, 2020 at 9:12

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answered Feb 12, 2015 at 17:17

Tivie

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1 Comment

Rafael Ruiz Muñoz Over a year ago

It was because of permissions! I copied all the involved folders (I'm making a copy) but there was one folder without the correct permissions (I created the folder instead copying all the properties). Thanks!

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