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

Jessica DiPonziano
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.a{fill-rule:evenodd;}techdegree
Jessica DiPonziano
Full Stack JavaScript Techdegree Student 20,782 Points
Posted by Jessica DiPonziano
.a{fill-rule:evenodd;}techdegree
Jessica DiPonziano
Full Stack JavaScript Techdegree Student 20,782 Points

My button isn't disappearing after the AJAX request is sent. Why?

The console tells me the error is

"Uncaught TypeError: Cannot read property 'style' of null"

As far as I can tell the code matches what is in the video. So why does the button not disappear. So confused. 

<!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');
    function sendAJAX() {
      xhr.send();
      document.getElementById("load").style.display = "none";
    }
  </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>
            <button id"load" onclick="sendAJAX()">Bring it!</button>
          </div>
          <div id="ajax">

          </div>
        </div>
      </div>
    </div>
  </div>
</body>
</html>
LaVaughn Haynes
LaVaughn Haynes
12,397 Points
LaVaughn Haynes
LaVaughn Haynes
12,397 Points

I edited your question to display the code properly

1 Answer

LaVaughn Haynes
LaVaughn Haynes
12,397 Points
LaVaughn Haynes
LaVaughn Haynes
12,397 Points 
<button id"load" onclick="sendAJAX()">Bring it!</button>

looks like your forgot your "=" sign. Should be id="load"

<button id="load" onclick="sendAJAX()">Bring it!</button>