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 trialPenny Gibson
10,519 PointsCan't get the request to send on part 2 of AJAX Basics "Finish the AJAX Request"
I've looked it up and can't figure out why this isn't working? Keeps asking if I've used the send.(); function.
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();
}
<!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
Penny Gibson
10,519 PointsThanks Brodey, It finally worked!!!
var request = new XMLHttpRequest(); request.onreadystatechange = function () { if (request.readyState === 4) { document.getElementById("footer").innerHTML = request.responseText; } }; request.open('GET', 'footer.html', sendAJAX()); function sendAJAX() { request.send(); }
Brodey Newman
10,962 PointsHey Penny!
You're so close! All you need to do is call your 'sendAJAX' function! :)
The code below should work.
sendAJAX();
Hope this helps!