Edward Edward -4 years ago 107
PHP Question

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

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?

Answer Source

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.

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