It seems like jQuery ".load()" is not working. What's the right alternative?

Question

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
  <link href='http://fonts.googleapis.com/css?family=Varela+Round' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" href="css/main.css">
  <script src="http://code.jquery.com/jqeury-1.11.0.min.js"></script>
  <script>

  function sendAJAX() {
    $('#ajax').load('sidebar.html');
  }
  </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>
          <button id="load" onclick="sendAJAX()" class="button">Bring it!</button>
          <div id="ajax">

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

This is my code which is exactly the same code on the video.

It seems that the $('#ajax').load('sidebar.html'); this part of the code is not working properly.

What would make this code working?

-AJAX Basics - Introducing jQuery-

Answer 1 · 2018-01-23T04:28:53Z

January 23, 2018 4:28am

There is nothing wrong with your code, it's just that you misspelled jquery. You wrote the cdn link as

<script src="http://code.jquery.com/jqeury-1.11.0.min.js"></script>

when it should have been

<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>

notice how you spelt jquery as jqeury

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HanJoon Kim

HanJoon Kim

It seems like jQuery ".load()" is not working. What's the right alternative?

1 Answer

Bjorn Beishline

Bjorn Beishline