Question

我在SO中找到了一些例子。但它们都不适合我。我想互相交换2 @Html.DropDownListFor的值。除了要显示的默认值之外，DropDownList都包含相同的值集。以下是我的代码。

<div class="currency_swap">
<p>
    <label>From</label>
    <div class="swap_dropdown">
        @Html.DropDownListFor(m => m.SourceCurrencyname, new SelectList(ViewBag.CurrencyNames, "Currency_Code", "Currency_Code"), "---- All ----")
    </div>
</p>
<div><input type="button" id="swap_up" value="&#x25B2" class="swap" /></div>
<p>
    <label>To</label>
    <div class="swap_dropdown">@Html.DropDownListFor(m => m.TargetCurrencyname, new SelectList(ViewBag.CurrencyNames, "Currency_Code", "Currency_Code"), "Agency Currency")</div>
</p>
</div>

<script type="text/javascript">
$(".swap").click(function () {
    var row = $(this).closest(".currency_swap");
    var start = row.find("swap_dropdown:first select");
    var end = row.find("swap_dropdown:last select");
    var temp = start.val();
    start.val(end.val());
    end.val(temp);
});
</script>

Answer 1

start和end变量未正确设置。选择器可能类似于以下内容。

var start = row.find(".swap_dropdown select:first");
var end = row.find(".swap_dropdown select:last");

其余代码有效，请参阅js-fiddle here。

如何用Jquery / JS交换2 @ Html.DropDownListFor

1 个答案: