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JavaScript AJAX Basics Programming AJAX Check for the correct ready state

Steven mudie
Steven mudie
5,523 Points
Posted by Steven mudie
Steven mudie
5,523 Points

this thing is broken

tells me to select the div with the sidebar ID, i do and keeps saying i need to do what i just did

app.js 
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
 if(xhr.readyState === 4 && xhr.statusText === "OK"){} ;
};
xhr.open('GET', 'sidebar.html');
xhr.send();
document.getElementById('sidebar');
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

Dimitar Dimitrov
Dimitar Dimitrov
11,800 Points
Dimitar Dimitrov
Dimitar Dimitrov
11,800 Points

document.getElementById('sidebar'); the selection code need to be in the conditional statement body. Like this : 

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