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

Anyone know why this won't work?

Any help appreciated. Thank you!

var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
request.open("GET", "footer.html");
 function sendAJAX() {
<!DOCTYPE html>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
  <script src="app.js"></script>
  <div id="main">
  <div id="footer"></div>

1 Answer

The last part of your code looks like this:

request.open("GET", "footer.html");
function sendAJAX() {

For the quiz, your request.send(); is correct, and is the only thing that is needed after the request.open(). Try removing the function sendAJAX(){} portion. The variable called "request" is already defined at the beginning of the code, so your last part of the code should be the following.

 request.open("GET", "footer.html");

@Mark Ihrig: I appreciate you explaining what was wrong more thoroughly than Dave's video did. I didn't understand the majority of the code he wrote because he didn't explain what he was using it for. So I didn't understand what function sendAJAX() { request.send();} meant or why it would be needed or not, so I included it just like Christopher did.