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JavaScript AJAX Basics Programming AJAX Check for the correct ready state

Posted by Steve Fau
Steve Fau
5,622 Points

document.querySelector('#sidebar') not accepted as a valid solution.

it's asking me to use document.getElementById('sidebar') but these should produce the same result afaik.

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
  if (xhr.readyState === 4 && xhr.status === 200) {
    document.querySelector('#sidebar')
  }
};
xhr.open('GET', 'sidebar.html');
xhr.send();
index.html 
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
  <script src="app.js"></script>
</head>
<body>
  <div id="main">
    <h1>AJAX!</h1>
  </div>
  <div id="sidebar"></div>
</body>
</html>

1 Answer

Lee Vaughn
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree seal-36
Lee Vaughn
Treehouse Teacher
Lee Vaughn
.a{fill-rule:evenodd;}techdegree seal-36
Lee Vaughn
Treehouse Teacher

Hi Steve Fau!

Nice work on coming up with that solution! There are a lot of different ways that you could go about selecting this element, and what you have in this example is one of them, but the challenge wants you to specifically use the getElementById() method.

I hope that helps!

Lee