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

mehmet odabas
mehmet odabas
2,120 Points
Posted by mehmet odabas
mehmet odabas
2,120 Points

There was an error with your code: TypeError: 'undefined' is not a function (evaluating 'request.Open('GET' , 'footer.ht

why request.Open('GET' , 'footer.html') is not working?

app.js 
var request = new XMLHttpRequest();
request.onreadystatechange = function () {

  if (request.readyState === 4) {  
    document.getElementById("footer").innerHTML = request.responseText;
  }
};

request.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>

2 Answers

Matthias J.
Matthias J.
20,355 Points
Matthias J.
Matthias J.
20,355 Points

you have to write ".open" in small letters :)

var request = new XMLHttpRequest();
request.onreadystatechange = function () {
    if (request.readyState === 4) {
        document.getElementById("footer").innerHTML = request.responseText;
    }
};
request.open('GET', 'footer.html');
mehmet odabas
mehmet odabas
2,120 Points
mehmet odabas
mehmet odabas
2,120 Points

thanks a lot. it works well but in the video it was typed as Open(..)