Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

JavaScript AJAX Basics (retiring) AJAX Concepts Finish the AJAX Request

Brooke Guy
Brooke Guy
8,179 Points

AJAX question

It's saying cant find variable xhr...

app.js
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};
xhr.open ('GET', 'footer.html');
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="footer"></div>
</body>
</html>

2 Answers

Zhaopeng Wang
seal-mask
PLUS
.a{fill-rule:evenodd;}techdegree seal-36
Zhaopeng Wang
Full Stack JavaScript Techdegree Graduate 32,210 Points
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};
request.open('GET','footer.html'); //in your code, you use var xhr, which is not defined.
//request.send();//this is for part 2
Steven Parker
Steven Parker
231,060 Points

Nothing named "xhr" has been created in this code. But the very top line creates "request", try calling the method on that instead.