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JavaScript AJAX Basics (retiring) AJAX Concepts A Simple AJAX Example

Posted by Matthew Aglibot
Matthew Aglibot
1,489 Points

I can't make the 'sidebar.html' appear.

No error's showing on Console.

<script> var xhr = new XMLHttpRequest();

xhr.onreadystatechange = function() {
  if (xhr.readyState === 4)
  {
    document.getElementById('ajax').innerHTML = xhr.responseText; 
  }

  xhr.open('GET', 'sidebar.html');
  xhr.send();

};

</script>

4 Answers

dragos busuioc
dragos busuioc
24,908 Points
dragos busuioc
dragos busuioc
24,908 Points 
xhr.onreadystatechange = function() { if (xhr.readyState === 4) { document.getElementById('ajax').innerHTML = xhr.responseText; 
} 
};
xhr.open('GET', 'sidebar.html'); //put it outside of onreadystatechange method 
xhr.send(); };//put it outside of onreadystatechange method
dragos busuioc
dragos busuioc
24,908 Points
dragos busuioc
dragos busuioc
24,908 Points

xhr.onreadystatechange = function() { if (xhr.readyState === 4) { document.getElementById('ajax').innerHTML = xhr.responseText; } xhr.open('GET', 'sidebar.html'); //put it outside of onreadystatechange method xhr.send(); };//put it outside of onreadystatechange method

T SL
PLUS
T SL
Courses Plus Student 378 Points
T SL
T SL
Courses Plus Student 378 Points

Even the examples in the folder doesn't work though. None of them show sidebar on Chrome on OSX

Christopher Gjelten
Christopher Gjelten
2,305 Points
Christopher Gjelten
Christopher Gjelten
2,305 Points

<script> var xhr = new XMLHttpRequest(); xhr.onreadystatechange = function () { if (xhr.readyState === 4 ) { document.getElementById('ajax').innerHTML = xhr.responseText; } }; xhr.open('GET', 'sidebar.html'); xhr.send(); </script>