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copied!<p><a href="https://stackoverflow.com/questions/4399617/python-os-getlogin-problem">This previous answer about Errno 22 when using <code>os.getlogin</code></a> should be enlightening:</p> <blockquote> <p>From the <code>os.getlogin()</code> <a href="http://docs.python.org/library/os.html#os.getlogin" rel="nofollow noreferrer">docs</a>: "Returns the user logged in to the controlling terminal of the process." Your script does not have a controlling terminal when run from cron. The docs go on to suggest: "For most purposes, it is more useful to use the environment variable <code>LOGNAME</code> to find out who the user is, or <code>pwd.getpwuid(os.getuid())[0]</code> to get the login name of the currently effective user id."</p> </blockquote> <p>You're calling the script via <code>passthru()</code> in PHP. PHP will be running as the web server user -- frequently <code>nobody</code> or <code>apache</code> -- and like cron, that user will also not be control a terminal.</p> <p>While you might be able to patch the code, you will probably be better served by filing a bug with the python-lastfm project with what you've uncovered. Unfortunately I don't know enough Python to help with that task, and also don't know enough about why the python-lastfm application wants to get the current user. It <em>looks</em> like it's just looking for a temporary file path and is trying to be clever about it.</p> <p>An alternative to patching might be setting environment variables for the script, but there isn't an apparent way to do this for <code>passthru</code>. Perhaps you can try using <a href="http://us2.php.net/manual/en/function.putenv.php" rel="nofollow noreferrer"><code>putenv('USER=myname')</code></a> before calling <code>passthru</code>, where 'myname' is your shell login name.</p>

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