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在MVC页面中获取第二个表单的输入

asp.net-mvc c#

Mister 832 · 技术社区 · 8 年前

我已经建立了一个页面,其中主控和详细信息(图片,aso)显示在一个页面上。但是,如果单击了第二个表单的submitt按钮,则如果第一个表单中的验证失败,则不会提交该表单。如果我更正这些值 HttpPostedFileBase uploadFile 为空。

页面如下所示:

@model app1.Models.MasterModel

@{
    ViewBag.Title = "Edit";
    Layout = "~/Views/Shared/_Layout.cshtml";
}

@using (Html.BeginForm(new { @class = "form-inline col-lg-12" }))
{
    @Html.AntiForgeryToken()

    <div>
        <h4>MasterModel</h4>
        <hr />
        @Html.ValidationSummary(true, "", new { @class = "text-danger" })
        @Html.HiddenFor(model => model.Id)


        <div class="row">
            @*Master properties*@

            <div class="col-md-4 col-lg-4">
                <div class="form-horizontal">
                    <div class="form-group">
                        @Html.LabelFor(model => model.Title, htmlAttributes: new { @class = "control-label col-md-3" })
                        <div class="col-md-8">
                            @Html.EditorFor(model => model.Title, new { htmlAttributes = new { @class = "form-control" } })
                            @Html.ValidationMessageFor(model => model.Title, "", new { @class = "text-danger" })
                        </div>
                    </div>

                 @* aso... *@

                </div>
            </div>


}


            @*  Master Details *@

            <div class="col-md-4 col-lg-4">



                @using (Html.BeginForm("NewPic", "Master", FormMethod.Post, new { enctype = "multipart/form-data" }))
                {
                    <input name="uploadFile" type="file" />
                    <input type="submit" value="Upload File" /> <!-- First Button, does not work -->

                    <div class="container-fluid">
                        @foreach (app1.Models.PicModel b in Model.Pics)
                        {

                            var base64 = Convert.ToBase64String(b.DbPic);
                            var imgSrc = String.Format("data:image/gif;base64,{0}", base64);


                <img src="@imgSrc" width="200" height="200" />
                        }
                    </div>


                    @Html.ActionLink("Upload", "NewPic", new { id = Model.Id }) <!-- Second Button, does not work either -->
                    <label class="control-label col-md-4 col-lg-4" for="Title">Picer</label>
                }
            </div>

        </div>

        <div>
            <div class="form-group">
                <div class="col-md-offset-2 col-md-12 col-lg-12">
                    <input type="submit" value="Save" class="btn btn-default" />
                </div>
            </div>
        </div>
    </div>


}

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




@section Scripts {
    @Scripts.Render("~/bundles/jqueryval")
}

控制器如下所示:

    public ActionResult NewPic(int id, HttpPostedFileBase uploadFile)
    {
         // uploadFile is null
    }

1 回复 | 直到 8 年前

Govinda Rajbhar 8 年前

你忘了把 [HttpPost] 在NewPic方法之前。所以 新建PIC 方法将被视为[HttpGet],因此它将不起作用。

[HttpPost]  
public ActionResult NewPic(int id, HttpPostedFileBase uploadFile)
{
     // uploadFile is null
}

并且还为这两个表单提供适当的Id,如下所示,以便在客户端验证时很容易使用这两个表单。

表格1

@using (Html.BeginForm(new {id = "Form1", @class = "form-inline col-lg-12" }))

表格2

@using (Html.BeginForm("NewPic", "Master", FormMethod.Post, new { id = "Form2", enctype = "multipart/form-data" }))

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