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brandonpe..

On mouseover, do X, on mouseout, do Y help?

in Using jQuery • 12 years ago

Hello all,

I'm new to this kind of scripting. I've managed to get my mouseover command working:

$('#saves_money').mouseover(function() {
$('.service_hide').hide();
document.getElementById('saves_money').innerHTML = '<img src="img/saves_money_over.jpg" />'
return false;
});

But I want it to undo itself when the cursor leaves... I tried:

$('#saves_money').mouseout(function() {
$('.service_hide').hide();
document.getElementById('saves_money').innerHTML = '<img src="img/saves_money.jpg" />'
return false;
});

But it didn't work. What's the syntax to make this happen?

Replies(6)

rminde

Re: On mouseover, do X, on mouseout, do Y help?

12 years ago

try this:

$('#saves_money').mouseover(function() {
$('.service_hide').hide();
$(this).find("img").attr("src", "img/saves_money_over.jpg");
});

brandonpe..

Re: On mouseover, do X, on mouseout, do Y help?

12 years ago

Isn't that also missing the on mouseout part? So by default in the HTML, image A is shown. On mouseover I want it to change to image B and when you take your curser off of it, I want it to turn to Image C.

rminde

Re: On mouseover, do X, on mouseout, do Y help?

12 years ago

for sure you have to write the code for mouseout too

brandonpe..

Re: On mouseover, do X, on mouseout, do Y help?

12 years ago

Well that's the part I'm missing. My mouseover code works fine. How do I get the mouseout part to work? When I change mouseover to mouseout and change the image, it does nothing. =/

rminde