Question

I am new to javascript and especially to ajax..And just trying to figure it out..

i wrote this code from one tutorial and can’t find what i did wrong. Here you can see it live

The error i am getting from Firebug: “TypeError: xmlhttp is undefined
[Break On This Error]

if (xmlhttp.readyState == 4){“

my code is

// JavaScript Document

var xmlhttp;
var url;

function ajaxFunction(){

if  (window.ActiveXObject){//if the window is InternetExplorer

    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");

    }else if(window.XMLHttpRequest){// if Window is Firefox etc..

        xmmlhttp= new XMLHttpRequest();

        }else{

            alert ("Get a New Browser")

            }

}//end of ajaxFunction()


function getInfo(){

    ajaxFunction();

    var entryInfo= document.getElementById("entry").value;

            function stateChanged(){
                if (xmlhttp.readyState == 4){

                    document.getElementById("results").innerHTML = xmlhttp.responseText;

                    }//if (xmlhttp.readyState == 4)

                }//end of function stateChanged()


url = "info.php?user="+entryInfo;
xmlhttp.onreadystateshange=stateChanged();
xmlhttp.open("GET",url,true);
xmlhttp.send(null);


    }// end of function getInfo

Answer 1

You have a typo here:

   xmmlhttp= new XMLHttpRequest();
     ^

Change to

 xmlhttp= new XMLHttpRequest();

Also as Michael pointed out, you have parenthesis when assigning your onreadystatechange function:

xmlhttp.onreadystateshange=stateChanged();
                                       ^ remove the ()

If you don’t remove the parenthesis, the stateChange() function will be called, and the return value will be given to xmlhttp.onreadystateshange which you don’t want.