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Start your free trialPaul May
3,432 PointsStubborn button
I managed to get everything working as expected, apart from when it came to removing the button. Can anyone spot my mistake?
<!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>
const xhr = new XMLHttpRequest();
xhr.onreadystatechange = () => {
if(xhr.readyState === 4){
document.querySelector('#ajax').innerHTML = xhr.responseText;
}
}
xhr.open('GET', 'sidebar.html');
const sendAJAX = () => {
xhr.send();
const button = document.querySelector('#load');
console.log(button);
/*
* Also tried
* button.remove
* and calling the hide method without setting the button variable
*/
button.style.visibility='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" onclick="sendAJAX()">Bring it!</button>
</div>
<ul id="ajax">
</ul>
</div>
</div>
</div>
</div>
</body>
</html>
I put the console.log in there hoping to find what I was doing wrong, but it returns the button element as expected.
<button id="load" onclick="sendAJAX()">Bring it!</button>
2 Answers
Harald N
15,843 PointsHi, I think the problem is that css visibility don't have a none property, however it has hidden. You can use style.display = 'none' or style.visibility = 'hidden' should also work.
Harald N
15,843 PointsNo problems Paul, I usually mix those up as well :D
Paul May
3,432 PointsPaul May
3,432 PointsThat worked perfectly I guess I got the two mixed up. Thanks for the reply