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Brooke Guy
8,179 PointsAJAX question
It's saying cant find variable xhr...
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
if (request.readyState === 4) {
document.getElementById("footer").innerHTML = request.responseText;
}
};
xhr.open ('GET', 'footer.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
Zhaopeng Wang
Full Stack JavaScript Techdegree Graduate 32,210 Pointsvar request = new XMLHttpRequest();
request.onreadystatechange = function () {
if (request.readyState === 4) {
document.getElementById("footer").innerHTML = request.responseText;
}
};
request.open('GET','footer.html'); //in your code, you use var xhr, which is not defined.
//request.send();//this is for part 2
Steven Parker
221,323 PointsNothing named "xhr" has been created in this code. But the very top line creates "request", try calling the method on that instead.