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

Brian Johnson
Brian Johnson
Front End Web Development Techdegree Graduate 18,997 Points

Can someone help me with this one?

Please help me!

app.js
var request = new XMLHttpRequest(); 
var xhr = 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>
eslam said
eslam said
Pro Student 6,733 Points
var request = new XMLHttpRequest(); 
request.open('GET', 'footer.html');
request.onreadystatechange = function () {
    if (request.readyState === 4) {
        document.getElementById("footer").innerHTML = request.responseText;
  }
}
request.send();

2 Answers

Steven Parker
Steven Parker
176,604 Points

The challenge has already created the XMLHttpRequest, you don't need to create another one.

Just apply the "open" method to the existing one named "request".