PHP: syntax error, unexpected 'else'

I got a syntax error when trying to set an username in my header, handling the exception when no first name and last name is seated, using the username to fill the label, surely is a stupid error. I'm using Quadodo Login Script in my project by the easily implementing.

Here is the code:

if(isset($qls->user_info['firstname'] { 
  $realname = "$qls->user_info['firstname'] . ' ' . $qls->user_info['lastname']"
} else { 
  $realname = "$qls->user_info['username']"
};

Thanks a lot in advantage. Andrea