我想实现创建使用$ .POST或$。阿贾克斯（INSERT）屏幕。

I want to achieve create (INSERT) screen using $.POST or $.AJAX.

注：code工作正常没有Ajax call..it已经在那里..现在我需要做的Ajax调用和保存而不回发。 我写了下面code：

Note: code is working fine without AJAX call..it is already there.. now i need to do ajax call and save without postback. I wrote below code:

在下面是code提交的单击事件：

On submit click event following is the code:

$.ajax(
      {
          url: '@Url.Action("CreateProduct","ProductManagement")',
          dataType: 'json',
          contentType: 'application/json; charset=utf-8',
          type: 'post',
          data:   $('frm').serialize(),
          success: function () { alert('s'); },
          error: function (e1, e2, e3) { alert(e2); alert(e1 + ' ' + e2 + ' ' + e3); alert('failure event'); }
      }
    );
});

在服务器端：

[HttpPost]
public JsonResult CreateProduct(FormCollection frm)
{
    manager.ProductManager m = new manager.ProductManager();
    // m.InsertProduct(new common.DTO.ProductDTO() { Id = id, ProductName = ProductName, Description = Description, Cost = cost, ProductTypeId = 5 });
    return Json("'result':'success'", JsonRequestBehavior.AllowGet);
}

但问题是，每一次，它调用的行动，但没有对 FRM 参数参数找到的数据。 我也试图保持的 - 视图模型 ProductViewModel VM 作为参数，但它并没有work..just给我空值。同时，在Ajax调用，这是不言而喻的成功，但。只是问题是没有张贴在控制器的行动。

Problem is , every time, it call the action but no data found on frm argument param. i also tried to keep as - View model ProductViewModel vm as argument but it did not work..just giving me null value. also, in ajax call, it goes in success but. just problem is nothing posted on controller's action.

HTML是如下：

  @using (Html.BeginForm("CreateProduct", "ProductManagement", FormMethod.Post, new { id = "frm" }))
    {
@Html.AntiForgeryToken()
@Html.ValidationSummary(true)

<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>

请指导我什么是错在这里。

Please guide me what is wrong here.

感谢您

推荐答案

有关id选择，你必须使用＃替换为标识。

For id Selector you have to use "#" with id.

这将工作：

$.ajax(
  {
      url: '@Url.Action("CreateProduct","ProductManagement")',
      dataType: 'json',
      contentType: 'application/json; charset=utf-8',
      type: 'post',
      data:   $('this').serialize(), OR  $('#frm').serialize(),   <-----------------
      success: function () { alert('s'); },
      error: function (e1, e2, e3) { alert(e2); alert(e1 + ' ' + e2 + ' ' + e3); alert('failure event'); }
  }
 );

或尝试这样的：

OR Try this :

var form = $('#frm');
$.ajax({
 cache: false,
 async: true,
 dataType: 'json',
 contentType: 'application/json; charset=utf-8',
 type: "POST",
 url: form.attr('action'),
 data: form.serialize(),
 success: function (data) {
 alert(data);
   }
 });