如何使用jquery在MVC中发布viewmodel [关闭]

Question

我没有使用过Form元素。我没有使用表格，因为，我不想回发..请指导我如何做ajax调用，从而能够将viewmodel传递给控制器​​的action方法获得$.ajax吗？我的表格如下所示：

HTML：

  @model comp.learn.data.Models.ProductViewModel

 @{
ViewBag.Title = "Create";
}

 <h2>Create</h2>


<fieldset>
    <legend>ProductViewModel</legend>
    <div id="CreateDiv">
        <div class="editor-label">
            @Html.LabelFor(model => model.ProductName)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.ProductName)
            @Html.ValidationMessageFor(model => model.ProductName)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.Cost)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.Cost)
            @Html.ValidationMessageFor(model => model.Cost)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.Description)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.Description)
            @Html.ValidationMessageFor(model => model.Description)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.ProductTypeId)
        </div>
        <div class="editor-field">
            @Html.DropDownList("ProductTypeId", "Choose item")
            @Html.ValidationMessageFor(model => model.ProductTypeId)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.ProductTypeName)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.ProductTypeName)
            @Html.ValidationMessageFor(model => model.ProductTypeName)
        </div>
    </div>
    <p>
        <input type="submit" value="Create" id="btnSubmit" />
    </p>

</fieldset>


<div>
  @Html.ActionLink("Back to List", "Index")
</div>

jQuery / JavaScript：

         $.ajax(
                {
                    url: '@Url.Action("CreateProduct","ProductManagement")',
                    dataType: 'json',
                    contentType: 'application/json; charset=utf-8',
                    type: 'post',
                    cache: false,
                    data: ///what should i write here ,
                    success: function (data) { alert('final'); },
                    error: function (f1, f2, f3) { alert(f3); }
                });

Answer 1

您应该手动从输入中收集数据，并构造与C＃模型类相对应的JSON对象。例如，如果您在操作方法中等待ProductViewModel对象，则可以遵循以下示例：

var myData = {
    productName: $('#ProductName').val(),
    cost: $('#Cost').val(),
    // .. and so on
};

$.ajax({
    data: JSON.stringify(myData),
    // .. the other ajax options
});

如果您具有表单元素，则更加容易。只需使用jQuery选择表单并调用serialize方法。数据将被编码为字符串以便提交。格式也将application/x-www- form-urlencoded; charset=UTF-8是$ .ajax默认值，您无需指定它。例：

var myData = $('#myFormId').serialize();
$.ajax({
    data: myData,
    contentType: 'application/x-www-form-urlencoded; charset=UTF-8',
    //..Other ajax options
});