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Peter Huang
5,427 Pointswhats wrong with my code?
doesent work
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>
1 Answer
Bella Bradbury
Front End Web Development Techdegree Graduate 32,790 PointsHi Peter!
The challenge is asking you to send the request. You've written a great function for this, but you'll need to call the function in order for it to actually send! To run your function you'll have to add sendAJAX() on the following line.
Another option is to take the send statement out of a function entirely and just place it below the open statement! In this case your code would look like:
...
request.open("GET", "footer.html");
request.send();
Either method above would use your existing code to have the request actually send!