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Daniel Toma
2,370 PointsOpen XMLHttpRequest with two arguments - code not working
I can't see what I'm doing wrong here. The task was to add the code to open the AJAX request using GET and pointing to 'footer.html'. I added the line of code below the IF statement, but I get the 'Bummer!' I don't know what else to try.
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
if (request.readyState === 4) {
document.getElementById("footer").innerHTML = request.responseText;
}
request.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>
3 Answers
Thayer Y
29,789 Pointsjust move the line outside the function, after the last curly bracket.
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
if (request.readyState === 4) {
document.getElementById("footer").innerHTML = request.responseText;
}
request.open('GET', 'footer.html');
};
request.open('GET', 'footer.html'); // move it here
Jon Schiedermayer
7,709 PointsI was stuck on exact same question. Thayer Y 's solution worked for me.
Daniel Toma
2,370 PointsThat was it! Thanks, Thayer Y!