Ubuntu – Run bash script from php

bashPHPscripts

I want to run a shell script within a php webpage to check some processes running on my system (Ubuntu 12.04). Googleing I've found about shell_exec() but I can't manage to run the script when loading php.

This is the php code I use.

$output = shell_exec('./dirlist.bash');
echo "<pre>$output</pre>";

and the piece of html I get

<pre></pre>

Also tried $output = shell_exec('sudo -u www-data ./dirlist.bash');

dirlist.bash does ls -l (just for testing script and shell_exec)

I think it is not a permission problem. Running directory grants write and execute privileges for all users (I should be careful here).

drwxrwxrwx  4 meteo meteo 4096 mar 11 15:20 RAMS

User www-data has been added to sudoers file in case it was needed

www-data ALL = (meteo) NOPASSWD: /home/meteo/www/RAMS/dirlist.bash

Thanks in advance

Best Answer

You probably need to chdir to the correct directory before calling the script. This way you can ensure what directory your script is "in" before calling the shell command.

$old_path = getcwd();
chdir('/my/path/');
$output = shell_exec('./script.sh var1 var2');
chdir($old_path);
echo "<pre>$output</pre>";

Related Solutions

Ubuntu – Run PHP script from command line

As long as you have php installed, you run a PHP file using

/usr/bin/php /path/to/php/file.php

Or if your $PATH is set up properly to include /usr/bin, then simply

php /path/to/php/file.php

You can check if PHP is installed, by running

which php

Ubuntu – Call and run php script from shell script

The problem here is that you are quoting the entiore command you are trying to run as a single variable. As a result, you're not running php with foo.php as an argument but instead are attempting to execute a file called php foo.php. Here's a simpler example to show you what I mean:

$ var1="echo "
$ var2="foo"
$ set -x ## debugging info
$ "$var1$var2"
+ 'echo foo'      ### The shell tries to execute both vars as a single command
bash: echo foo: command not found

$ "$var1" "$var2"
+ 'echo ' foo     ### The shell tries to execute 'echo ' (echo with a space)
bash: echo foo: command not found

So, the right way is to remove the space and quote each variable separately:

$ var1="echo"
$ var2="foo"
$ "$var1" "$var2"

If you do that though, you'll hit the next error. The . is the source command. That tries to read a shell script and execute it in the current session. You are giving it a php script instead. That won't work, you need to execute it, not source it.

Finally, always avoid using CAPITAL variable names. The shell's reserved variables are capitalized so it's a good idea to always use lower case variable names for your scripts.

Putting all this together (with a few other minor improvements), what you want is something like:

#!/bin/sh

list="/path/to/my/site/dir"
config="/usr/bin/php"

for i in "$list"
do
    "$config" "$i"/test.php
done

Best Answer

Related Solutions

Ubuntu – Run PHP script from command line

Ubuntu – Call and run php script from shell script

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