Reputation: 37
I have 4 image buttons. I wrote some JavaScript code for changing the color of image on mouseenter. Also on mouseleave, it will back to the original state as shown below.
<script language="javascript" type="text/javascript">
function mouseOverImage(Id) {
document.getElementById(Id).src = "Images/Yellow.png";
}
function mouseOutImage(Id) {
document.getElementById(Id).src = "Images/TestImage.png";
}
</script>
If I click on button1, I have to change the color to yellow and mouse leave event could not be fired on this situvation.
If I click on button2
, button1
should back to original state.
Can anyone help?
Upvotes: 1
Views: 1602
Reputation: 522
Its easier if you use jQuery, Here's some sample code
$('#button1').on('click', function () {
//button 1 click event goes here
$(this).css("background-image", "url(Images/Yellow.png)");
});
$('#button1').hover(
function () {
//button 1 hover event
},
function () {
//button 1 hover out event
});
$('#button2').on('click', function () {
//button 2 click event goes here
$('#button1').css("background-image", "url(Images/Original.png)");
});
Upvotes: 2
Reputation: 911
you can try,
$(document).on(
{
mouseenter: function () {
},
mouseleave: function () {
}
}, '.chat-button');
chat-button is -> button2 and button1 client id
jquery 1.9.x
Upvotes: 0