JavaScript AJAX Basics (retiring) jQuery and AJAX Handle AJAX failures with jQuery

Pitrov Secondary
Pitrov Secondary
5,119 Points

What is wrong here?

Can someone help me with this one?

app.js
$.get("missing.html", function(data) {
  $("#footer").html(data);
}).fail(
let jqXHR = new jqXHR;
);
index.html
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>AJAX with JavaScript</title>
</head>
<body>
  <div id="main">
    <h1>AJAX with jQuery</h1>
  </div>
  <div id="footer"></div>
  <script src="jquery.js"></script>
  <script src="app.js"></script>
</body>
</html>

2 Answers

You need to pass a function to the fail method with the parameter. Currently you are just setting a variable.

$.get("missing.html", function(data) {
  $("#footer").html(data);
}).fail(function(jqXHR){
alert(jqXHR.statusText);
});