PHP: Why isn't exec() returning output?

Question

I'm writing a PHP script to be used to check for network connections with Linux shell command ping calling it with PHP's exec():

<?php


// Bad IP domain for testing.
$domain_bad = "lksjdflksjdf.com";

$ip_address = $domain_bad;

exec("ping -c 1 $domain_bad", $output, $return_var);

var_dump($return_var);
echo "return_var is: $return_var" . "\n";
var_dump($output);


exit;
?>

I'm not getting the output for the error message from ping in $output which is what I'm expecting:

$ php try.php
ping: unknown host lksjdflksjdf.com
int(2)
return_var is: 2
array(0) {
}

If the domain is a good domain, such as yahoo.com, then $output has the output from ping in an array. But if it's an error such as 'ping: unknown host lksjdflksjdf.com' it doesn't get returned to the $output array.

Why is this happening and is there a better method to do this?

Community · Accepted Answer · 2017-05-23 10:31:14Z

156

You should redirect stderr to stdout.

To do that, change your exec() call like this:

exec("ping -c 1 $domain_bad 2>&1", $output, $return_var);

More info about 2>&1 meaning here.

edited May 23, 2017 at 10:31

CommunityBot

11 silver badge

answered May 21, 2013 at 8:12

Hariprasad

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1 Comment

D T Over a year ago

add 2>&1 but output still is array empty, if execute from cmd , it output 2 values

Phạm Tuấn Anh · Accepted Answer · 2020-01-02 03:46:00Z

10

If the answer above could not solve your problem, maybe exec() is disabled. You can try to check php.ini file at disable_functions line.

edited Jan 2, 2020 at 3:46

answered Mar 7, 2017 at 6:30

Phạm Tuấn Anh

1641 silver badge12 bronze badges

1 Comment

makozaki Over a year ago

Good hint. Should be disable_functions instead of disabled_functions

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