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 trial
Alonzo Delk
6,991 PointsAJAX request
I'm in need of some? My request code to send is not working.
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";
}
<!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
Full Stack JavaScript Techdegree Student 19,560 PointsHey 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
Full Stack JavaScript Techdegree Student 19,560 PointsFor 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.
Alonzo Delk
6,991 PointsHey Bramyn thanks for the help. For some reason it's still not passing but I'll figure it out. Happy Code
Alonzo Delk
6,991 PointsYes 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!