I have 2 multi select boxes like as in this link. /

$(function(){
    $("#button1").click(function(){
        $("#list1 > option:selected").each(function(){
            $(this).remove().appendTo("#list2");
        });
    });

    $("#button2").click(function(){
        $("#list2 > option:selected").each(function(){
            $(this).remove().appendTo("#list1");
        });
    });
});

But When i add from one first select box to second select box it is working fine for me.But again when i add from second select box to first select box they are appending to last of first select box.But what i want is i need to add they must be added in the place where they deleted.

Thanks in advance...

I have 2 multi select boxes like as in this link. http://jsfiddle/bdMAF/38/

$(function(){
    $("#button1").click(function(){
        $("#list1 > option:selected").each(function(){
            $(this).remove().appendTo("#list2");
        });
    });

    $("#button2").click(function(){
        $("#list2 > option:selected").each(function(){
            $(this).remove().appendTo("#list1");
        });
    });
});

But When i add from one first select box to second select box it is working fine for me.But again when i add from second select box to first select box they are appending to last of first select box.But what i want is i need to add they must be added in the place where they deleted.

Thanks in advance...

• javascript
• jquery

• Did you find a better method? – Edorka Commented Nov 5, 2013 at 11:19

3 Answers 3

Maybe you can simply set and remove the attribute "disabled" but it will leave a blank space in the options. Note that with this method you will need to clone the option the first time.

The other solution whould be to add the content as usual but applying a sort() function before

function byValue(a, b) {
    return a.value > b.value ? 1 : -1;
};

function rearrangeList(list) {
    $(list).find("option").sort(byValue).appendTo(list);
}

$(function () {
    $("#button1").click(function () {
        $("#list1 > option:selected").each(function () {
            $(this).remove().appendTo("#list2");
            rearrangeList("#list2");
        });
    });

    $("#button2").click(function () {
        $("#list2 > option:selected").each(function () {
            $(this).remove().appendTo("#list1");
            rearrangeList("#list1");
        });
    });
});

You can try at this fiddle

You need to keep track of some sort of index, I think in your case you can use the value of each option. (but you could use a dedicated data-* attribute if you need to)

With that value you can then search the current list and see where it should fit in. Loop the options and check for a value greater than the one you are moving. If you find one then insert it before that, if you don't fine one then insert at the end.

This should do it:

$("#button2").click(function(){
    $("#list2 > option:selected").each(function(){
        var item = $(this).remove();
        var match = null;
        $("#list1").find("option").each(function(){
            if($(this).attr("value") > item.attr("value")){
                match = $(this);
                return false;
            }
        });
        if(match)
            item.insertBefore(match);
        else
           $("#list1").append(item);
    });
});

You can apply the same for the reverse too.

Here is a working example

After adding the options back to #list1, a simple sort() will do the rest. For that we need to add a parison function to it based on its value.

$(function () {
    $("#button1").click(function () {
        $("#list1 > option:selected").each(function () {
            $(this).remove().appendTo("#list2");
        });
    });

    $("#button2").click(function () {
        $("#list2 > option:selected").each(function () {
            $(this).remove().appendTo("#list1");
            var opts = $('#list1 option').get();

            $('#list1 ').html(opts.sort(optionSort));
        });
    });

    function optionSort(a, b) {
        return $(a).val() > $(b).val();
    }
});

Check out this JSFiddle

You can also sort using text() instead of val() by changing it in the optionSort().