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Rails 3 の CoffeeScript のビルドを使用して、項目をある選択リストから別の選択リストに移動したり戻したりしています。これは、次のコードで機能します。

$ ->
   $('#add').click (event) ->
     $('#geo_country_exclude option:selected').remove().appendTo('#geo_country_include');  

   $('#remove').click (event) ->
     $('#geo_country_include option:selected').remove().appendTo('#geo_country_exclude');

選択リストのコード:

<%= select :geo, :country_exclude, @countries, {}, {:multiple => true, :size => 15} %>
<%= select :geo, :country_include, @countries, {}, {:multiple => true, :size => 15} %>

問題は、リストの最後に新しい項目を追加することです。リストをもう一度アルファベット順に並べ替えたいと思います。どうすればいいですか？ありがとう。

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1 に答える 1

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Basically, you need to pull the <option>s out of the target <select>, add the option to that array, sort it by hand, and then replace the target <select>s <option>s with the sorted list. jQuery doesn't help that much with this sort of thing but the low level DOM version isn't terribly difficult, something like this:

exc = $('#geo_country_exclude')[0]
inc = $('#geo_country_include')[0]

by_option_value = (a, b) ->
    return +1 if(a.value > b.value)
    return -1 if(a.value < b.value)
    return  0

mv = (from, to) ->
    # to.options looks like an array and smells like an array but
    # it isn't an array so we employ the standard arrayification trick.
    opts = [].slice.call(to.options)
    # Move the <option>
    opts.push(from.options[from.selectedIndex])
    from.remove(from.selectedIndex)
    # Sort and replace.
    to.remove(0) for i in [ 0 ... to.options.length ]
    to.add(opt) for opt in opts.sort(by_option_value)

$('#add'   ).click (event) -> mv(exc, inc)
$('#remove').click (event) -> mv(inc, exc)

You might need to tweak the by_option_value comparison function and you might want to adjust the selected items in the <select>s after the move but you can sort that you easily enough.

Demo: http://jsfiddle.net/ambiguous/qRw3K/

If you need to deal with multi-selects then a minor modification to mv will do the trick:

mv = (from, to) ->
    opts = [].slice.call(to.options)
    while from.selectedIndex >= 0
        opts.push(from.options[from.selectedIndex])
        from.remove(from.selectedIndex)
    to.remove(0) for i in [ 0 ... to.options.length ]
    to.add(opt) for opt in opts.sort(by_option_value)

You just need to wrap the opts.push and from.remove in a little while loop.

Demo: http://jsfiddle.net/ambiguous/qRw3K/1/

于 2012-08-18T17:55:45.560 に答える