PHP StdErr after Excec()

Question

In PHP i am executing a command with exec(), and it returns if successful an URL;

$url = exec('report');

However, I want to check STD_ERR, if something went wrong. How would I read the stream? I want to use php://stderr, but not sure how to use it.

Thanks, Martin

Answer 1

According to this, it's

 $url = exec("$com 2>&1", $out, $err); 

has always worked for me.

Answer 2

If you want to execute a command, and get both stderr and stdout, not "merged", a solution would probably to use proc_open, which provides a great level of control over the command that's being executed -- including a way to pipe stdin/stdout/stderr.

And here is an example : let's consider we have this shell-script, in test.sh, which writes to both stderr and stdout :

#!/bin/bash

echo 'this is on stdout';
echo 'this is on stdout too';

echo 'this is on stderr' >&2;
echo 'this is on stderr too' >&2;

Now, let's code some PHP, in temp.php -- first, we initialize the i/o descriptors :

$descriptorspec = array(
   0 => array("pipe", "r"),  // stdin
   1 => array("pipe", "w"),  // stdout
   2 => array("pipe", "w"),  // stderr
);

And, then, execute the test.sh command, using those descriptors, in the current directory, and saying the i/o should be from/to $pipes :

$process = proc_open('./test.sh', $descriptorspec, $pipes, dirname(__FILE__), null);

We can now read from the two output pipes :

$stdout = stream_get_contents($pipes[1]);
fclose($pipes[1]);

$stderr = stream_get_contents($pipes[2]);
fclose($pipes[2]);

And, if we output the content of those two variables :

echo "stdout : \n";
var_dump($stdout);

echo "stderr :\n";
var_dump($stderr);

We get the following output when executing the temp.php script :

$ php ./temp.php
stdout :
string(40) "this is on stdout
this is on stdout too
"
stderr :
string(40) "this is on stderr
this is on stderr too
"

Hope this helps :-)