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Home/ Questions/Q 7024833
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T23:53:39+00:00 2026-05-27T23:53:39+00:00

I’m trying to create a checklist that will sort checked items to the top

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I’m trying to create a checklist that will sort checked items to the top of the list, and when an item is unchecked, it will move below the checked items. Beyond the checked/unchecked sort, items should stay in their original order, regardless of what order they are checked in.

In other words, a five item list should always look like this:
* two
* three
one
four
five

not this:
*three
*two
one
four
five

I’ve seen examples of how to move clicked items to a specific position in the list, but nothing that takes into account the state of other items in the list.

Ideally I’d also like to animate the transition. Also, is there a way to remember the checked state across multiple pages?

This is the HTML I’m starting with:

<form>
<ul>
<li><label for="one"><input type="checkbox" id="one" />One</label></li>
<li><label for="two"><input type="checkbox" id="two" />Two</label></li>
<li><label for="three"><input type="checkbox" id="three" />Three</label></li>
<li><label for="four"><input type="checkbox" id="four" />Four</label></li>
<li><label for="five"><input type="checkbox" id="five" />Five</label></li>
</ul>
</form>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T23:53:39+00:00Added an answer on May 27, 2026 at 11:53 pm

    Select the original list to save the order. When one of the checkboxes is clicked, iterate over it, appending the checked items first and the unchecked items second:

    var list = $("ul"),
        origOrder = list.children();
    
    list.on("click", ":checkbox", function() {
        var i, checked = document.createDocumentFragment(),
            unchecked = document.createDocumentFragment();
        for (i = 0; i < origOrder.length; i++) {
            if (origOrder[i].getElementsByTagName("input")[0].checked) {
                checked.appendChild(origOrder[i]);
            } else {
                unchecked.appendChild(origOrder[i]);
            }
        }
        list.append(checked).append(unchecked);
    });
    

    Demo: http://jsfiddle.net/scSYV/2

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