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

Nina Kozlova
Nina Kozlova
11,602 Points
Posted by 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.

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if(xhr.readyState ===4 && xhr.status === 200){}
};
xhr.open('GET', 'sidebar.html');
xhr.send();
document.getElementById('sidebar');
index.html 
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
  <script src="app.js"></script>
</head>
<body>
  <div id="main">
    <h1>AJAX!</h1>
  </div>
  <div id="sidebar"></div>
</body>
</html>

2 Answers

James Anwyl
.a{fill-rule:evenodd;}techdegree seal-36
James Anwyl
Full Stack JavaScript Techdegree Graduate 49,960 Points

Hi,

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){
  document.getElementById('sidebar');
}
};
xhr.open('GET', 'sidebar.html');
xhr.send();

Hope this helps :)