成功登录后转到另一个页面

登录后我正试图转到另一个页面，那么如何在使用Ajax和php成功登录后转到另一页面？当ajax向php发送请求，然后得到类似的响应时（如果响应是（'true'）或（'false'），则此响应取决于检查用户名和密码），所以我不知道如何在ajax中处理此响应，以将其作为传递到下一页的条件。请帮忙。

var HttPRequest = false;
   function doCallAjax(Mode) {
      HttPRequest = false;
      if (window.XMLHttpRequest) { // Mozilla, Safari,...
         HttPRequest = new XMLHttpRequest();
         if (HttPRequest.overrideMimeType) {
            HttPRequest.overrideMimeType('text/html');
         }
      } else if (window.ActiveXObject) { // IE
         try {
            HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
         } catch (e) {
            try {
               HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e) {}
         }
      }
      if (!HttPRequest) {
         alert('Cannot create XMLHTTP instance');
         return false;
      }
      var url = 'PHP_Login.php';

      var pmeters = "tEmail=" + encodeURI( document.getElementById("Email").value) +
                    "&tpassword=" + encodeURI( document.getElementById("password").value ) +
                    "&tMode=" + Mode;

        HttPRequest.open('POST',url,true);

        HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        HttPRequest.setRequestHeader("Content-length", pmeters.length);
        HttPRequest.setRequestHeader("Connection", "close");
        HttPRequest.send(pmeters);

        HttPRequest.onreadystatechange = function()
        {
             if(HttPRequest.readyState == 3)  // Loading Request
              {
               document.getElementById("mySpan").innerHTML = "Now is Loading...";
              }
             if(HttPRequest.readyState == 4) // Return Request
              {
               document.getElementById("mySpan").innerHTML = HttPRequest.responseText;
               document.getElementById("Email").value = '';
               document.getElementById("password").value = '';
           }
            }
        }