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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
221,323 Points

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