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

Posted by Terri Valentine
Terri Valentine
23,034 Points

Add the code to send the AJAX request.

```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(); send.AJAX(); }

What am I missing? It keeps saying "Bummer. Did you call send() on the request object?

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();
  send.AJAX();
}
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

Jazz Jones
Jazz Jones
8,535 Points
Jazz Jones
Jazz Jones
8,535 Points 
request.open('GET', 'footer.html');
request.send();

^^ is all you need to complete the challenge. I'm not sure what you were trying to do with your sendAJAX function as there isn't a .AJAX method in vanilla javascript. Remember AJAX is a technique, not an actual built in method and requires 4 things:

  1. XMLHttpRequest Object
  2. Callback function
  3. Open method to open a request
  4. Send method to send the request.

Hope this helps :)

Terri Valentine
Terri Valentine
23,034 Points

Thank you Jazz Jones. I should have taken a break from the problem and came back with a fresh set of eyes. Lol. Thank you for posting a reminder of the AJAX technique requirements.

Leandro Botella Penalva
Leandro Botella Penalva
17,618 Points
Leandro Botella Penalva
Leandro Botella Penalva
17,618 Points

Hi Terri,

I don't know where did you get that send.AJAX() function since it doesn't exist. Also, you don't need to create a function. Call the request send method after the request is opened like 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();
Terri Valentine
Terri Valentine
23,034 Points

Thank you Leandro Penalva. I had been working on a solution for a couple of hours. I am a newbie. I was trying just about everything. I guess I should have taken a break. Lol.