Reputation: 6755
<div id="menu_1"><img src="replacement"/></div>
<div id="menu_2"><img src="replacement"/></div>
<div id="menu_3"><img src="replacement"/></div>
<div id="menu_1_hover"><img src="onhover_userreplacement_img"/></div>
<div id="menu_2_hover"><img src="onhover_userreplacement_img"/></div>
<div id="menu_3_hover"><img src="onhover_userreplacement_img"/></div>
How would I on hovering over menu_1, menu_2 and menu_3 replace the images with the respective images located in menu_1_hover, menu_2_hover, menu_3_hover ?
Upvotes: 0
Views: 754
Reputation: 6055
$("#menu_1").hover(function () {
$(this).html($("div #"+$(this).attr("id")+"_hover").html());
}
That should work for "menu_1" you should set a class for all of the elements though so you only have to put the .hover thing once. Not tested. :)
Upvotes: 0
Reputation: 9382
$(document).ready(function() {
$("#menu_1").hover(function() {
var id = $(this).attr("id");
var src = $(this).children("img").attr("src");
$("#" + id + "_hover").children("img").attr("src", src);
});
});
A working example: http://jsfiddle.net/gtWCY/
To add a single hover event listener I'd suggest adding a class like "hover" to all of the menu_'s.
Upvotes: 0
Reputation: 9242
Better to use CSS instead of JQuery, but if you must do it in JQuery, I can post the answer, just inform me. :)
Upvotes: 1