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JavaScript AJAX Basics AJAX Concepts Finish the AJAX Request

AJAX request

I'm in need of some? My request code to send is not working.

app.js
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};
request.open('GET', 'footer.html');
    function sendAJAX() {
     request.send();
     document.getElementById('load').style.display = "none";
    }
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>

3 Answers

Bramyn Payne
Bramyn Payne
19,589 Points

Hey Alonzo,

Try this:

var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};
request.open('GET', 'footer.html');
request.send();

You did not need to put it into a function for this. Hopefully this helps!

Bramyn Payne
Bramyn Payne
19,589 Points

For part 2 of the challenge, you need to call send() on request.

request.send();

Like you have it, but it should not be in your request.open(). It should be below.

Hey Bramyn thanks for the help. For some reason it's still not passing but I'll figure it out. Happy Code

Yes I was trying to add it to the function as well so it wouldn't work. But after placing the request.send() code under the brackets it worked. Thanks!