Hi all,
Here's what I would like to do. I have 2 identical lists that I would like to have change color on hover of a specific item on both lists simultaneously.
Example:
List One
List Item 1
List Item 2
List Item 3
List Two
List Item 1
List Item 2
List Item 3
So if you were to hover over List One's first item it would highlight List Two's first item also.
Any help on this would save my life thanks very much. V
Just off the top of my head, you could use jquery to select a specific tag in each list and apply a style to it. Maybe use the .addClass method to add a css style to the same items in the array of tags for both select lists.
Sample code:
<style>
.active { color: red; font-weight: bold; }
</style>
<body>
<ul id="list1">
<li>foo</li>
<li>bar</li>
</ul>
<ul id="list2">
<li>foo</li>
<li>bar</li>
</ul>
<script>
(function() {
var list1 = document.getElementById('list1'),
list2 = document.getElementById('list2');
function setActive(target, b) {
if(!target.tagName.toLowerCase() === 'li') return;
// determine position in list:
for(var i = 0; target = target.previousSibling; ++i);
list1.childNodes[i].className = b ? 'active' : '';
list2.childNodes[i].className = b ? 'active' : '';
}
// mouseover/mouseout are bubbling, so add the listeners to parents:
list1.onmouseover = list2.onmouseover = function(event) {
setActive((event && event.target) || window.event.srcElement, true);
};
list1.onmouseout = list2.onmouseout = function(event) {
setActive((event && event.target) || window.event.srcElement, false);
};
})();
</script>
HTML
<ul id="list1">
<li>item 1</li>
<li>item 2</li>
</ul>
<ul id="list2">
<li>item 1</li>
<li>item 2</li>
</ul>
jQuery
$("#list1 li").hover(
function(){
var ind = $("#list1 li").index($(this));
$("#list2 li:eq(" + ind + ")").addClass("highlight");
$(this).addClass("highlight");
},
function(){
var ind = $("#list1 li").index($(this));
$("#list2 li:eq(" + ind + ")").removeClass("highlight");
$(this).removeClass("highlight");
}
);
The highlight class will have the style that you wish to apply to both lists.