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JavaScript AJAX Basics Programming AJAX Check for the correct ready state

Posted by Christopher Evans
Christopher Evans
9,896 Points

Checking for ready state - code should work

My code identical to the code from the video, but with the conditional for "xhr status 200" portion removed. Why wouldn't this code work?

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function () {
  if (xhr.readyState === 4 {
     alert("Alles goed");
  });
};
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

Sean T. Unwin
Sean T. Unwin
28,690 Points
Sean T. Unwin
Sean T. Unwin
28,690 Points

The closing round bracket is in the wrong place. It is currently after the if statement's closing curly bracket, whereas it should be after the conditional, before the opening curly bracket.

  if (xhr.readyState === 4 {
     alert("Alles goed");
  });
// ^ Move this to after the `4` and remove the semi-colon (`;`).