sh: 2: 1: not found - pass multiple php arguments to bash script

Question

I have a simple bash script which i call from my php code to find out the version of my apache and nginx.

$webroot = getcwd();

function get_version($name)
{
    global $webroot;

    switch ($name)
    {
        case "apache":
            $path   = shell_exec("whereis apachectl | awk '{ print $2 }'");
            $version = shell_exec("sudo $webroot/scripts/get_version $path 1 2>&1");
            break;
        case "nginx":
            $path = shell_exec("whereis nginx | awk '{ print $2 }'");
            $version = shell_exec("sudo $webroot/scripts/get_version $path 2 2>&1");
        default:
            echo "error";
    }

    return $version;
}

As you can see i call my bash script with two arguments passed. The path and a integer number which i use in my bash script:

#!/bin/bash

_x=""
_programm=$1
_nr=$2

if [ "$_nr" -eq "1" ] ; then
    _x=$($_programm -v 2>/dev/null | grep -i 'version' | awk -F/ '{ print $4 }')
elif [ "$_nr" -eq "2" ] ; then
    _x=$($_programm -v 2>&1 | awk -F/ '{ print $2 }')
fi

cd $(pwd)
echo $_x

Output of function:

get_version("apache");     OUTPUT:     sh: 2: 1: not found
get_version("nginx");      OUTPUT:     sh: 2: 2: not found

But if i execute the bash script in the terminal, then it works and i get the version number as output, i tried it both with user root and www-data, both worked. The bash script is also entered in the visudo file and has execute rights, the user of the script is www-data.

./get_version /usr/sbin/apachectl 1     OUTPUT: 2.2.2
./get_version /usr/sbin/nginx 2         OUTPUT: 1.3

Can someone please explain why it does work in terminal but not in php?

Do you mean like this: print \$2 ? I tried it but then i get syntax errors. — Black
– Black, Commented Feb 1, 2016 at 9:48

Black · Accepted Answer · 2016-02-01 12:22:38Z

2

i found the problem and the solution. The command whereis in my php switch statement wrote a space character to the path variable for some unknown reason, so it did not worked because of it. I used rtrim on my $path variable to fix it.

    case "apache":
        $path   = shell_exec("whereis apachectl | awk '{ print $2 }'");
        $path   = rtrim($path);
        ...

edited Feb 1, 2016 at 12:22

answered Feb 1, 2016 at 10:50

Black

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1 Comment

user1934428 Over a year ago

Does it work now? Note also, that the error message you received, was not produced by bash, but by sh.

maxhb · Accepted Answer · 2016-02-01 09:53:04Z

1

You have to escape $ if you are using it inside double quotes in php or switch to single quotes:

...
$path   = shell_exec('whereis apachectl | awk \'{ print $2 }\'');
...
$path = shell_exec('whereis nginx | awk \'{ print $2 }\'');

answered Feb 1, 2016 at 9:53

maxhb

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7 Comments

Black Over a year ago

Hmm. I used your code but unfortunately i still get the same error. I also think it makes no different to my code.

maxhb Over a year ago

Please add debug output for $path and $version string and post in your question. echo $path . "\n" . $version; before break statement.

Black Over a year ago

I get /usr/sbin/apachectl + sh: 2: 1: not found and /usr/sbin/nginx + sh: 2: 2: not found

maxhb Over a year ago

If I set $webroot to /my/web/root I get sudo /my/web/root/scripts/get_version whereis apachectl | awk '{ print $2 }' 1 2>&1 which seems to be fine...

maxhb Over a year ago

Just a remark: You can simplify your script by using which apachectl instead of whereis apachectl. First will just return the path to the executable, no additional info.

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