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JavaScript AJAX Basics (retiring) Programming AJAX Check for the correct ready state

Nina Kozlova
Nina Kozlova
11,602 Points

So what is wrong with this?

I am pretty sure this is how it's supposed to be done. What's more - even the instructions tell me that's how it's done. Yet the code doesn't go through and I'm stuck on task 3 of 4.

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if(xhr.readyState ===4 && xhr.status === 200){}
xhr.open('GET', 'sidebar.html');
<!DOCTYPE html>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
  <script src="app.js"></script>
  <div id="main">
  <div id="sidebar"></div>

2 Answers


You need to place the document.getElementById('sidebar') inside the if statement, as we want to run this code when the server responds with status 200.

For example:

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if(xhr.readyState === 4 && xhr.status === 200){
xhr.open('GET', 'sidebar.html');

Hope this helps :)