Question

我正在尝试上传文件，但是在保存文件时会引发操作错误，即未设置对象引用，并且值也为NULL。

与传递和保存该类相同，但仍会引发错误。为什么它为空？它绑定到相同的视图模型。

@model VAILCertificates.UI.ViewModels.AddInspectionReportViewModel

@using (Html.BeginForm("Create", "InspectionReport", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
     <div class="form-group">
          @Html.LabelFor(model => model.File, htmlAttributes: new { @class = "control-label col-md-2" })
          <div class="col-md-5">
               @Html.TextBoxFor(model => model.File, new { @class = "form-control", type = "file" })
               @Html.ValidationMessageFor(model => model.File, "", new { @class = "text-danger" })
          </div>
     </div>
     <button class="btn btn-success btn-outline btn-block" type="submit"><i class="fa fa-save" style="font-size:medium"></i>  Add</button>
}

操作：

public ActionResult Create(AddInspectionReportViewModel model, HttpPostedFileBase FileName)   
    {
         string FilePath = Server.MapPath("~/Downloads/Certificates/");
                            model.File.SaveAs(FilePath + Path.GetFileName(model.File.FileName));
         model.InspectionReport.FileName = Path.GetFileName(model.File.FileName);

        InspectionReportDAL.AddInspectionReport(model.InspectionReport);
        return RedirectToAction("Index");
    }

Viewmodel：

public class AddInspectionReportViewModel
{
     public HttpPostedFile File { get; set; }
     public InspectionReport InspectionReport { get; set; }
}

班级：

public class InspectionReport
    {
        //General Details
        public int InspectionReportID { get; set; }


        public string FileName { get; set; }
    }

更新：内部错误是

The parameter conversion from type 'System.Web.HttpPostedFileWrapper' to type 'System.Web.HttpPostedFile' failed because no type converter can convert between these types.

Answer 1

HttpPostedFileBase FileName

model.File

该方法期望的参数名称是FileName。

编辑此@Html.TextBoxFor(model => model.File, new { @class = "form-control", type = "file" })

到

@Html.TextBoxFor(model => model.File.FileName , new { @class = "form-control", type = "file" })

根据评论，我将尽可能执行以下操作。

如果我是你，我将删除@Html.TextBoxFor(model => model.File, new { @class = "form-control", type = "file" })并替换为：

<input type="file" name="FileName" id="fileUpload or whatever" accept="//Whatever file you want to accept">

在当前状态下，您永远不会传递任何参数到HttpPostedFileBase参数，仅传递模型，因此它始终为null。

Answer 2

将viewModel更改为：

 public class AddInspectionReportViewModel
            {
                [DataType(DataType.Upload)]
                public HttpPostedFileBase File { get; set; }
                public InspectionReport InspectionReport { get; set; }
            }

将视图更改为：

@using (Html.BeginForm("Create", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    <div class="form-group">
        @Html.LabelFor(model => model.File, htmlAttributes: new { @class = "control-label col-md-2" })
        <div class="col-md-5">
            @Html.TextBoxFor(m => m.File, new { type = "file" })
            @Html.ValidationMessageFor(model => model.File, "", new { @class = "text-danger" })
        </div>
    </div>
    <button class="btn btn-success btn-outline btn-block" type="submit"><i class="fa fa-save" style="font-size:medium"></i>  Add</button>
}

Answer 3

将HttpPostedFile更改为HttpPostedFileBase可行。

 public class AddInspectionReportViewModel
    {
        public HttpPostedFileBase File { get; set; }
        public InspectionReport InspectionReport { get; set; }
    }

为什么HttpPostedFile在视图中的操作方法中为null？

3 个答案: