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

Roger Fleenor
Roger Fleenor
647 Points
Posted by 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