jQuery交换ids

Question

当我点击#inactive id时，#active-page应该会消失，而#inactive-page应该会出现。此外，#active id应该成为#inactive id，反之亦然。我使用了将id的=>#inactive-page-swap与#inactive-page的品质进行交换，将#active-swap与#active的品质进行交换。

问题是-当我单击#inactive id时，#active-page消失了，但#inactive-page没有出现。

<script>
$(document).ready(function(){
    $("#inactive").click(function(){
        $(#inactive).attr("id","active-swap");
        $("#active").attr("id","inactive");
        $("#active-swap").attr("id","active");
        $("#active-page").attr("id","inactive-page-swap",function(){
            $("#inactive-page").attr("id","active-page");
            $("#inactive-page-swap").attr("id","inactive-page");
        });
    });
});
</script>

请帮帮我！

Answer 1

试试这个。

$(document).ready(function(){
    $("#inactive").click(function(){
        var id = $(this).attr('id');
        inactive = $('#inactive');
        active = $('#active');
        swap = $('#active-swap');
        inactive.attr('id', 'active-swap');
        active.attr('id', 'inactive');
        swap.attr('id', 'active');
        inactive = $('#active-page');
        inactivepage = $('#inactive-page');
        inactiveswap = $('#inactive-page-swap');
        $(inactive).attr("id","inactive-page-swap");
        $(inactivepage).attr("id","active-page");
        $(inactiveswap).attr("id","inactive-page");
    });
});

查看此jsFiddle example