code: function log() {
    if ($.fn.ajaxSubmit.debug) {
      var msg = '[jquery.form] ' + Array.prototype.join.call(arguments,'');
      if (window.console && window.console.log) {
        window.console.log(msg);
      }
      else if (window.opera && window.opera.postError) {
        window.opera.postError(msg);
      }
    }
  }
docs: helper fn for console logging set $.fn.ajaxSubmit.debug to true to enable debug logging