PHP StdErr after Exec()

PhpExecStderr

Php Problem Overview


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

$url = exec('report');

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

Php Solutions


Solution 1 - Php

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
"

Solution 2 - Php

A little function that might be helpful:

function my_shell_exec($cmd, &$stdout=null, &$stderr=null) {
    $proc = proc_open($cmd,[
        1 => ['pipe','w'],
        2 => ['pipe','w'],
    ],$pipes);
    $stdout = stream_get_contents($pipes[1]);
    fclose($pipes[1]);
    $stderr = stream_get_contents($pipes[2]);
    fclose($pipes[2]);
    return proc_close($proc);
}

The exit code is returned and STDOUT and STDERR are reference params if you need them.

Solution 3 - Php

The short way to do such a things with exec is to return the exit code ( status of the command )

Note that I am trying to list a non-exists directory /non-dir/

exec('ls /non-dir/', $out, $retval);
var_dump($retval);

Output > ls: cannot access '/non-dir/': No such file or directory
> int(2)

Normally in unix-based system most of successful statuses codes is ( 0 ) so you can check your $retval to know the status of the command.

to dismiss the error from listing an invalid path ls: cannot access '/non-dir/': No such file or directory you can redirect your stderr to null

exec('ls /non-dir/ 2>/dev/null', $out, $retval);
var_dump($retval);

this will output :

> int(2)

also if you need the error string to use it in any scenario you may redirect your stderr to the stdout.

exec('ls /non-dir/ 2>&1', $out, $retval);
print_r($out);
var_dump($retval);

this will output the following:

Array
(
    [0] => ls: cannot access '/non-dir/': No such file or directory
)
int(2)

Solution 4 - Php

Another way to get unmerged stdout/stderr.

$pp_name = "/tmp/pp_test";
@unlink($pp_name);
posix_mkfifo($pp_name, 0777);
$pp = fopen($pp_name, "r+");
stream_set_blocking($pp, FALSE);
exec("wget -O - http://www.youtube.com 2>$pp_name", $r_stdout);
$r_stderr = stream_get_contents($pp);
var_dump($r_stderr);
fclose($pp);
unlink($pp_name);

If you want to ignore stdout and get only stderr, you can try this:

exec("wget -O - http://www.youtube.com 2>&1 >/dev/null", $r_stderr);

Solution 5 - Php

exec("{$command} 2>&1"
    ,$output
    ,$exitCode
    );
  • 2>&1 redirects stderr to stdout for consistent success / fail behaviour.

  • $exitCode determines $command completion status.

  • $output contains all output associated with $exitCode.

Solution 6 - Php

Slightly ugly but good enough. Put the stderr into a temp file and read it back.

$tmp = tempnam("/tmp", "ERR_");
exec('report 2> ' . escapeshellarg($tmp), $stdout, $retcode);
$stderr = file_get_contents($tmp);
unlink($tmp);
if ($retcode == 0)
{
    // good
    $url = $stdout[0];
} else {
    // bad
    error_log("FAIL: $stderr");
}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionmartinView Question on Stackoverflow
Solution 1 - PhpPascal MARTINView Answer on Stackoverflow
Solution 2 - PhpmpenView Answer on Stackoverflow
Solution 3 - PhphassanView Answer on Stackoverflow
Solution 4 - PhpLi-chih WuView Answer on Stackoverflow
Solution 5 - PhpPanchoView Answer on Stackoverflow
Solution 6 - PhpTelView Answer on Stackoverflow