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

Posted by Christopher Bryant
Christopher Bryant
220 Points

Anyone know why this won't work?

Any help appreciated. Thank you!

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

Mark Ihrig
Mark Ihrig
19,966 Points
Mark Ihrig
Mark Ihrig
19,966 Points

The last part of your code looks like this:

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

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");
        request.send();
A X
A X
12,842 Points

@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.