下面的代碼生成 5 個下拉列表。@{for (int i = 0; i < 5; i++) { <tr> <td> @Html.Editor("[" + i + "].QNo", new { htmlAttributes = new { @class = "form-control ", @type = "text", @placeholder = " QNo", @required = "", @id = "txtQNo", @style = "width:60px;" } }) </td> <td> </td> <td> @Html.DropDownList("[" + i + "].Question", new SelectList(string.Empty, "Value", "Text"), "Select Question", new { @class = "form-control ", @id = "Question", @style = "width:900px;" })</td> </tr> } }我正在嘗試使用通過下面的 ajax 調(diào)用收到的一堆值來填充上面的 5 個下拉菜單$("#ReflectionType").on("change", function (event) { $.ajax({ type: "post", url: "/Question/GetQuestions", data: { TypeId: $('#ReflectionType').val() }, datatype: "json", traditional: true, success: function (data) { debugger; $.each(data, function (index, value) { var markup = ''; $("#Question").append('<option value="' + value.Question + '">' + value.Question + '</option>'); }); } });上面的代碼片段只更新了一個下拉列表（第一個下拉菜單），它應(yīng)該更新所有五個下拉列表。
1 回答
函數(shù)式編程
TA貢獻1807條經(jīng)驗 獲得超9個贊
@{
for (int i = 0; i < 5; i++) { <tr>
<td> @Html.Editor("[" + i + "].QNo", new { htmlAttributes = new { @class = "form-control ", @type = "text",
@placeholder = " QNo", @required = "", @id = "txtQNo", @style = "width:60px;" } })</td>
<td> </td>
<td> @Html.DropDownList("[" + i + "].Question", new SelectList(string.Empty,"Value", "Text"), "Select Question",
new { @class = "form-control ", @id = "Question"+i, @style = "width:900px;" })</td>
</tr>
}
}
這將生成唯一的 id，如下所示 Question0、Question1、Question2、Question3、Question4
$("#ReflectionType").on("change", function (event) {
$.ajax({
type: "post",
url: "/Question/GetQuestions",
data: { TypeId: $('#ReflectionType').val() },
datatype: "json",
traditional: true,
success: function (data) {
debugger;
$.each(data, function (index, value) {
var markup = '';
for(let j = 0; j < 5; j++){
$("#Question"+j).append('<option value="' + value.Question + '">' + value.Question + '</option>');
}
});
}
});
如我所見，您的循環(huán)正在運行 5 次迭代，您可以以相同的方式運行它并附加來自 ajax 調(diào)用的數(shù)據(jù)?；蛘吣梢允褂?starts with selector 如下
$('[id^=Question]').each(function(index,element){
$(element).append('<option value="' + value.Question + '">' + value.Question + '</option>');
})
希望這能解決您的問題?？鞓肪幋a！
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