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JavaScript AJAX Basics AJAX Concepts Finish the AJAX Request

Posted by William Ray Noble
William Ray Noble
22,190 Points

Error message says that request.send(); is wrong.

Hello, I've been over my code several times as well as the code I wrote for the previous exercise in Workspaces. I cannot find the error that I am making, thank you in advance!

app.js 
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('footer').style.display = "none";
}
index.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>

1 Answer

Blake Larson
Blake Larson
13,014 Points
Blake Larson
Blake Larson
13,014 Points

try this

var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};

request.open('GET', 'footer.html', true);
request.send();

live example --> https://www.w3schools.com/xml/tryit.asp?filename=tryxml_httprequest

William Ray Noble
William Ray Noble
22,190 Points

Thank you, that worked.