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JavaScript AJAX Basics AJAX Concepts A Simple AJAX Example

Turandeep Munday
Turandeep Munday
6,047 Points

Can't set ajax element even though when i run the code `document.getElementById('ajax').innerHTML = xhr.responseText;`

My code copies the teachers code, however when i run the code in the browser it doesn't work.

after running in browser when i manually set the elements using the console document.getElementById('ajax').innerHTML = xhr.responseText; - this works and i can see that the response returns a 200 from XHR/Fetch portion of the console

But I am unsure why when i recieve this response the text isn't updating the responseTExt

code below

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <link href='http://fonts.googleapis.com/css?family=Varela+Round' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" href="css/main.css">
  <title>AJAX with JavaScript</title>
  <script>
  var xhr = new XMLHttpRequest();

  xhr.onreadystatechange = function () {
    if (xhr.readyState === 4) {
      document.getElementById('ajax').innerHTML = xhr.responseText;
    }
  };
  xhr.open('GET', 'sidebar.html');
  xhr.send();
  </script>
</head>
<body>
  <div class="grid-container centered">
    <div class="grid-100">
      <div class="contained">
        <div class="grid-100">
          <h1>Bring on the AJAX</h1>
          <ul id="ajax">

          </ul>
        </div>
      </div>
    </div>
  </div>
</body>
</html>
Steven Parker
Steven Parker
231,268 Points

A better way to share workspace code (particularly when multiple files are involved) is to make a snapshot of your workspace and post the link to it. This allows others to replicate the issue and provide a more complete and accurate answer.

1 Answer

josephweiss2
josephweiss2
7,409 Points

Try this:

xhr.onreadystatechange = function () {
    if (this.readyState == 4 && this.status == 200) {
      document.getElementById('ajax').innerHTML = xhr.responseText;
    }
  };

Let me know if it still doesn't work