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JavaScript AJAX Basics AJAX Concepts A Simple AJAX Example

Stubborn button

I managed to get everything working as expected, apart from when it came to removing the button. Can anyone spot my mistake?

<!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>
      const xhr = new XMLHttpRequest();
      xhr.onreadystatechange = () => {
        if(xhr.readyState === 4){
          document.querySelector('#ajax').innerHTML = xhr.responseText;
        }
      }
      xhr.open('GET', 'sidebar.html');
      const sendAJAX = () => {
        xhr.send();
        const button = document.querySelector('#load');
        console.log(button);
        /* 
        * Also tried  
        * button.remove
        * and calling the hide method without setting the button variable 
        */
        button.style.visibility='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>
          <ul id="ajax">
          </ul>
        </div>
      </div>
    </div>
  </div>
</body>
</html>

I put the console.log in there hoping to find what I was doing wrong, but it returns the button element as expected.

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

2 Answers

Harald N
Harald N
15,843 Points

Hi, I think the problem is that css visibility don't have a none property, however it has hidden. You can use style.display = 'none' or style.visibility = 'hidden' should also work.

That worked perfectly I guess I got the two mixed up. Thanks for the reply

Harald N
Harald N
15,843 Points

No problems Paul, I usually mix those up as well :D