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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
19,589 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
19,589 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!