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

Posted by Brian Johnson
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Brian Johnson
Front End Web Development Techdegree Graduate 20,818 Points

Can someone help me with this one?

Please help me!

app.js 
var request = new XMLHttpRequest(); 
var xhr = new XMLHttpRequest(); 
request.onreadystatechange = function () {
  if (request.readyState === 4) {
    document.getElementById("footer").innerHTML = request.responseText;
  }
};
xhr.open("GET", "footer.html");
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>
eslam said
eslam said
Courses Plus Student 6,734 Points
eslam said
eslam said
Courses Plus Student 6,734 Points 
var request = new XMLHttpRequest(); 
request.open('GET', 'footer.html');
request.onreadystatechange = function () {
    if (request.readyState === 4) {
        document.getElementById("footer").innerHTML = request.responseText;
  }
}
request.send();

2 Answers

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

The challenge has already created the XMLHttpRequest, you don't need to create another one.

Just apply the "open" method to the existing one named "request".

Brian Johnson
seal-mask
PLUS
.a{fill-rule:evenodd;}techdegree seal-36
Brian Johnson
Front End Web Development Techdegree Graduate 20,818 Points
Brian Johnson
.a{fill-rule:evenodd;}techdegree seal-36
Brian Johnson
Front End Web Development Techdegree Graduate 20,818 Points

Thanks Steve!