Question

我传入一个文件名，然后我想要request.files但它一直返回null。这是代码：

 @using (Html.BeginForm("AddTechNote", "Ticket", FormMethod.Post))
 <div class="col-md-10">
       <input type="file" id="fileToUpload" name="fileToUpload" />
       <span class="field-validation-error" id="spanfile"></span>
 </div>

并在控制器中：

public ActionResult AddTechNote(TicketView ticketReturn, string Note, bool PublicNote, string fileToUpload, string CopyIntoEmail)
{
        HttpPostedFileBase file = Request.Files[fileToUpload];
        string _fileName = null;

        if (file != null && file.ContentLength > 0)
        {
               _fileName = new FileController().UploadFile(file, "Tickets", ticketReturn.TicketNumber.ToString());
        }

视图不是强类型的（至少不是这个模型）。输入字段位于表单内。

Answer 1

您需要拥有string fileToUpload并在后续代码中使用HttpPostedFileBase fileToUpload，而不是fileToUpload，如下所示。

public ActionResult AddTechNote(TicketView ticketReturn, string Note, bool PublicNote, HttpPostedFileBase fileToUpload, string CopyIntoEmail)
{
        string _fileName = null;

        if (fileToUpload != null && fileToUpload.ContentLength > 0)
        {
               _fileName = new FileController().UploadFile(fileToUpload, "Tickets", ticketReturn.TicketNumber.ToString());
        }
  ...........
}

另外，请确保为multipart/form-data设置了BeginForm，例如 -

@using (Html.BeginForm("Index", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))

Request.Files在mvc中按名称命名

1 个答案: