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JavaScript AJAX Basics Programming AJAX Check for the correct ready state

Posted by Sardar Ahmed Khan
Sardar Ahmed Khan
308 Points

code for handling response. I am getting the syntax error?

I am handling the onreadystatechange event.

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if(xhr.readystate===4){
  document.getElementByID('').InnerHtml = xhr.response;
};
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

Torben Korb
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PLUS
.a{fill-rule:evenodd;}techdegree seal-36
Torben Korb
Front End Web Development Techdegree Graduate 91,394 Points
Torben Korb
.a{fill-rule:evenodd;}techdegree seal-36
Torben Korb
Front End Web Development Techdegree Graduate 91,394 Points

Hi Sardar,

seems like you're missing a closing bracket for the onreadystagechange handler function. Should look like here:

xhr.onreadystatechange = function () {
    if (xhr.readystate === 4) {
        document.getElementByID('').InnerHtml = xhr.response;
    }
};

Hope this helps. Happy coding!

Sardar Ahmed Khan
Sardar Ahmed Khan
308 Points

Thanks for help! I got the error solved myself.