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

shirshah sahel
shirshah sahel
10,034 Points

I try to log in XMLHttpRequest in console but it gives me error.

<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');
    xrh.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>
          <div id="ajax">

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

https://w.trhou.se/yiguz59oem.

Jonathan Mitten
Jonathan Mitten
Pro Student 11,172 Points

Hi shirshah, I don't see any console.log() in your code nor in the attached workspace. Where are you trying to log the response object? Please provide the code exactly as it is when it fails. Thanks!

3 Answers

Anders Blom
Anders Blom
8,689 Points

Hi shirshah, at the end of your script-tag, you seem to have a small typo. It should be ´xhr.send()rather than ´xrh.send().

That small correction should make your AJAX request send properly, and fill up the #ajax-div.

shirshah sahel
shirshah sahel
10,034 Points

Thanks Mitten, it was a miss spilling issue.

shirshah sahel
shirshah sahel
10,034 Points

Thanks Mr Blom, That is exactly what it was. after the correction of mis spilling it worked.