JavaScript AJAX Basics (retiring) AJAX Concepts A Simple AJAX Example

Posted by baderbouta
baderbouta
7,802 Points

Stuck, I've checked everything and It's still not working.

Heya'.

I can't seem to wrap my head around to why this is not working? Can anyone please help me with what the heck is going wrong in this code?

<!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>

  // Step 1 // - Create XMLHTTP Request object.
  var xhr = new XMLHttpRequest();

  // Step 2 // - Create callback function. 
  //this is the code that is going to run when the server respons to your request.
  xhr.onreadystatechange = function () {
    if (xhr.readyState === 4) {
      document.getElementById('ajax').innerHTML = xhr.responseText;
    }
  };

  // Step 3 // - Open a request.
  xhr.open('GET', 'sidebar.html');


  //Step 4 // - Send Request 
  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>
Adam Beer
Adam Beer
11,270 Points
Adam Beer
Adam Beer
11,270 Points

This is looks good. Maybe you don't saved the file before check

1 Answer

Sean T. Unwin
MOD
Sean T. Unwin
Treehouse Moderator 28,388 Points
Sean T. Unwin
Sean T. Unwin
Treehouse Moderator 28,388 Points

Missing the button that when pressed calls the the function which xhr.send(); is wrapped in. The function is missing as well.

Check the video from the 9:50 mark.

baderbouta
baderbouta
7,802 Points

Fixed it thanks!