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JavaScript AJAX Basics AJAX Concepts A Simple AJAX Example

Posted by Joshua Ouma
Joshua Ouma
20,865 Points

Why doesn't my button work?

I have followed the steps in the video correctly, yet my button still doesn't work as expected. why?

Joshua Ouma
Joshua Ouma
20,865 Points 
 <!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <link href='//fonts.googleapis.com/css?family=Varela+Round' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" href="css/main.css">
  <title>AJAX with JavaScript</title>
  <script>
    var xhr = new XMLHttpRequest();
    xhr.onreadystatechange = function(){
      if(xhr.readyState === 4){
        document.getElementById("ajax").innerHTML = xhr.responseText;
      }
    };
    xhr.open("GET", "sidebar.html");
    function sendAJAX(){
      xhr.send();
      document.getElementById("load").style.display = "none";
    }
  </script>
</head>
<body>
  <div class="grid-container centered">
    <div class="grid-100">
      <div class="contained">
        <div class="grid-100">
          <div class="heading">
            <h1>Bring on the AJAX</h1>
            <button id="load" onlick="sendAJAX();">Bring it!</button>
          </div>
          <div id="ajax">

          </div>
        </div>
      </div>
    </div>
  </div>
</body>
</html>
Joshua Ouma
Joshua Ouma
20,865 Points

I have included my code

Joshua Ouma
Joshua Ouma
20,865 Points

Yes, I have included everything. Actually it works perfectly without the button. But when I add the button, clicking the button doesn't display the expected text

Joshua Ouma
Joshua Ouma
20,865 Points

Yes

Daniel Turato
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Daniel Turato
Java Web Development Techdegree Graduate 30,124 Points
Daniel Turato
.a{fill-rule:evenodd;}techdegree seal-36
Daniel Turato
Java Web Development Techdegree Graduate 30,124 Points

Can you take a snapshot of your workspace? You can do this by clicking the camera button on the top right of your workspace. Once done, copy and paste the link of the snapshot here so I can see

Nelson J
Nelson J
7,411 Points

I am having the same problem

2 Answers

Kevin Gates
Kevin Gates
15,052 Points
Kevin Gates
Kevin Gates
15,052 Points

You put onlick() instead of onclick(). You're missing the "c" in "click".

horacinis
horacinis
4,824 Points
horacinis
horacinis
4,824 Points

<button id="load" onlick="sendAJAX();">

Yeah, you put "onlick" instead of "onclick" but I see another mistake, you put a semicolon at the end of "sendAJAX();", remove that semicolon so it its "sendAJAX()", without the semicolon.