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JavaScript AJAX Basics Programming AJAX Check for the correct ready state

Now that the server has responded with the data, you need to add it to the page. First select the div with the ID of 'si

what am I getting wrong here

app.js
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState === 4 && xhr.status === 200) {
};
};
xhr.open('GET', 'sidebar.html');
xhr.send()
 console.log(xhr.responseText)
document.getElementById('sidebar');
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

Heidi Fryzell
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MOD
.a{fill-rule:evenodd;}techdegree seal-36
Heidi Fryzell
Front End Web Development Treehouse Moderator 25,178 Points

Hi Jedidiah,

The code you wrote is correct, it is just where you put it that is the issue 🙂. It needs to be inside of the if conditional statement because you don't want it to run until the server has sent back its response and received the ok status.

So if you move your code where you select the sidebar inside the if statement it should work.

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState === 4 && xhr.status === 200) {
  document.getElementById('sidebar');
};
};
xhr.open('GET', 'sidebar.html');
xhr.send()
console.log(xhr.responseText)

Happy coding! Heidi