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JavaScript AJAX Basics (retiring) Programming AJAX Check for the correct ready state

Posted by Justin Henderson
Justin Henderson
13,739 Points

Add a conditional statement inside the onreadystatechange event handler that tests to make sure the server has sent back

I believe that I only have part of the puzzle here, not sure what to do next. Any contribution would be greatly appreciated.

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() { if (xhr.readystate===4&& xhr.status===200){

};
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>

5 Answers

Kelly von Borstel
Kelly von Borstel
28,880 Points
Kelly von Borstel
Kelly von Borstel
28,880 Points

The code is missing closing bracket after function, and readystate should be readyState, with uppercase S -- like this:

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() { 
  if (xhr.readyState === 4 && xhr.status === 200) {
  };
};
xhr.open('GET', 'sidebar.html');
xhr.send();
Justin Henderson
Justin Henderson
13,739 Points

Hello Kelly,

Thank you for this response, it was very helpful! I wasn't paying close enough attention to the event handler. Thanks again!

Justin

Tom Nguyen
Tom Nguyen
33,501 Points
Tom Nguyen
Tom Nguyen
33,501 Points

This is what I used to pass the challenge

app.js 
var xhr = new XMLHttpRequest();
  xhr.onreadystatechange = function() {
    if (xhr.readyState === 4){  // 4 means the readystate has send everything that it will send
      if (xhr.status === 200){
        document.getElementById('sidebar').innerHTML = xhr.responseText;
      }
    }
};
xhr.open('GET', 'sidebar.html');
xhr.send();
Volodymyr Rusnyk
Volodymyr Rusnyk
9,657 Points
Volodymyr Rusnyk
Volodymyr Rusnyk
9,657 Points

ar xhr = new XMLHttpRequest(); xhr.onreadystatechange = function() { if (xhr.readyState === 4 && xhr.status === 200) { }; xhr.open('GET', 'sidebar.html'); xhr.send();

I wrote the same code and it doesn’t work. It shows: Bummer: There was an error with your code: SyntaxError: Parse error What is wrong with code?

ellie adam
ellie adam
26,377 Points
ellie adam
ellie adam
26,377 Points

var xhr = new XMLHttpRequest(); xhr.onreadystatechange = function() { if (xhr.readyState === 4 && xhr.status === 200) {

};

xhr.open('GET', 'sidebar.html'); xhr.send(); };

Raynaldo Joseph
Raynaldo Joseph
8,187 Points
Raynaldo Joseph
Raynaldo Joseph
8,187 Points

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