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JavaScript AJAX Basics (retiring) AJAX Concepts Finish the AJAX Request

Posted by Erica Rottman
Erica Rottman
2,602 Points

add code to send ajax request

Eror that im not calling request.send(), but i am

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();
}
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

ve
ve
5,048 Points
ve
ve
5,048 Points

Actually, you didn't really send the request. What you did is you created a function that sends the request, but if that function isn't called, the request isn't sent. Place your request.send() outside of your function or call the function at the end of your code using sendAJAX(); , both ways should properly send your request.