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

Roger Fleenor
Roger Fleenor
647 Points

why is this wrong

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

app.js
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function(){
  if(xhr.readyState === 4 && xhr.status === 200){
    document.getElementById('sidebar').innerHTML() = xhr.responseText;
  }
};
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