Reputation: 265
I have a hidden div that gets displayed when a button is clicked that shows a spinner, but if you hit the back button after leaving the page the spinner is still visible. I think the javascript needs the $(document).ready(function() {
but I'm not sure how to implement it. I don't want the spinner visible if the user presses the back button.
This is what I have so far. The Javascript
<script type="text/javascript">
function showDiv() {
document.getElementById('spinner').style.display = "block";
}
</script>
The Div and link used to show the div
<div id="spinner" style="display:none;" class="spinner" ><img src="images/loader2.gif" alt="Loading"/></div>
<a class="index_buttons" href="search_members.php" onclick="showDiv()">Search Members </a>
Upvotes: 2
Views: 598
Reputation: 32949
use document ready as follows:
$(document).ready( function() {
hideSpinner(); // Initiates the hide function when document is ready.
});
function hideSpinner() {
$('#spinner').hide();
}
Upvotes: 0
Reputation: 70728
To hide the spinner on page load do:
$(document).ready(function() {
$("#spinner").hide();
}
Upvotes: 2
Reputation: 18848
Since you're tagged this as jQuery..
/* Calls this function on ready */
$(function() {
/* Hide the spinner whenever the page loads */
$("#spinner").hide();
;
$(".index_buttons").click(function() { $("#spinner").show(); });
});
Now, everytime a button with the class index_buttons
is clicked, the button will be visible. If you want to toggle the spinner, you might want to look at toggle()
.
Upvotes: 0