Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
204 views
in Technique[技术] by (71.8m points)

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: $return_var" . "
";
var_dump($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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...