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

Floyd Orr
Floyd Orr
11,723 Points
Posted by Floyd Orr
Floyd Orr
11,723 Points

code not working

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <link href='//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">
          <div class="heading">
            <h1>Bring on the AJAX</h1>
          </div>
          <ul id="ajax">

          </ul>
        </div>
      </div>
    </div>
  </div>
</body>
</html>

2 Answers

Kaetlyn McCafferty
Kaetlyn McCafferty
12,193 Points
Kaetlyn McCafferty
Kaetlyn McCafferty
12,193 Points

In your open() method, you have an extra parenthesis after 'GET'. It should look like this:

xhr.open('GET', 'sidebar.html');