exec() - Apache says file doesn't exists

[lobobr]

Greetings.

I'm trying to execute a simple C program in a Linux server and return it's output to the browser.

The only way I could did this was with the backtick and configuring PHP to run in unsafe mode. It works, but I still wanna do it with exec(). And I want PHP in safe mode again, so please somebody help me here.

This is what I've tried:

<?php

$results = array();

exec('./program_name', $results); // I've used ./ because the program is in same directory of the script

for ($n = 0; $n < count($results); $n++)
{
echo $results[$n]."<br>";
}
?>

When I try this Apache generates on it's error log a line saying file not found.

I've tried everything you can poossibily imagine: run other commands, like exec ('/bin/ls'... , replacing the simple quotes with double quotes, etc, etc...

I've even did a getenv to see the correct path of the program and it's there.

I've also tried absolute path, so this is not the problem.

If I do:

$a = `./program_name';

with safe mode off, it works.

What's happening here? Please somone help me.

Thanks in advance,

Er Galvão Abbott

[Anon]

The section of the manual on safe mode says that if the program isn't in a directory defined by the safe_mode_include_dir configuration variable, it won't run. And that variable is blank by default. The same restriction applies to popen(), and the backtick operator is disabled. I guess that's why they call it safe mode.

[lobobr]

Thanks a lot for your reply. I'll take a look.

Best regards,