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JavaScript AJAX Basics (retiring) AJAX Concepts Finish the AJAX Request

Posted by Alisson Matheus Fachini Soares
Alisson Matheus Fachini Soares
Courses Plus Student 685 Points

What's wrong

var request = new XMLHttpRequest(); request.onreadystatechange = function () { if (request.readyState === 4) { document.getElementById("footer").innerHTML = request.responseText; request.open('GET', 'footer.html', true); request.send(); } };

app.js 
var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
    request.open('GET', 'footer.html', true);
    request.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="footer">

  </div>
</body>
</html>

1 Answer

Jennifer Nordell
seal-mask
STAFF
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Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

The open and send requests needs to be outside of the callback function. Take a look:

var request = new XMLHttpRequest();
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};

request.open('GET', 'footer.html', true);
request.send();

You're doing great! The code isn't really wrong... it's just a little misplaced :)

Alisson Matheus Fachini Soares
Alisson Matheus Fachini Soares
Courses Plus Student 685 Points

Thank you