shell_exec in PHP returns empty string

Question

shell_exec and exec are not returning any content. I can't figure out what's wrong.

Here's some code:

echo 'test: ';
$output = shell_exec('whoami');
var_export($output, TRUE);
echo PHP_EOL . '<br>' . PHP_EOL;

And here's the source of the output

test 2: 
<br>

I do not have control over the host, but I believe they're running SuPHP. According to phpinfo, safe_mode is off. Running whoami from SSH outputs the expected value.

I'm at a loss. Any idea how to debug this?

lanzz · Accepted Answer · 2012-07-10 21:43:50Z

5

You're never printing the $output variable. The var_export() call returns the content of the variable when you call it with a true second parameter, it does not print it directly.

answered Jul 10, 2012 at 21:43

lanzz

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2 Comments

sharoz Over a year ago

I changed it to echo var_export($output, TRUE);. Now it outputs NULL. Any ideas?

drew010 Over a year ago

@sharoz Try shell_exec('/usr/bin/whoami'); instead, and make sure /usr/bin is the correct path.

paulsm4 · Accepted Answer · 2012-07-10 21:43:46Z

0

If you want the output from a shell command read back into PHP, you're probably going to need popen(). For example:

if( ($fp = popen("some shell command", "r")) ) {
    while( !feof($fp) ) {
        echo fread($fp, 1024);
        flush(); // input will be buffered
    }
    fclose($fp);
}

answered Jul 10, 2012 at 21:43

paulsm4

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