MVC 3文件上传和模型绑定


91

我有一个可以上载的表单,但是我想传递数据库的模型信息,以使用其他名称保存文件。

这是我的剃刀视图:

@model CertispecWeb.Models.Container

@{
  ViewBag.Title = "AddDocuments";
}

<h2>AddDocuments</h2>

@Model.ContainerNo

@using (Html.BeginForm("Uploadfile", "Containers", FormMethod.Post, 
            new { enctype = "multipart/form-data" }))
{
    <input type='file' name='file' id='file' />
    <input type="submit" value="submit" />
}

这是我的控制器:

[HttpPost]
public ActionResult Uploadfile(Container containers, HttpPostedFileBase file)
{
     if (file.ContentLength > 0)
     {
        var fileName = Path.GetFileName(file.FileName);
        var path = Path.Combine(Server.MapPath("~/App_Data/Uploads"),
                       containers.ContainerNo);
        file.SaveAs(path);
     }

     return RedirectToAction("Index");
}

模型信息不会传递到控制器。我读到我可能需要更新模型,我该怎么做?


2
请参阅stackoverflow.com/questions/9411563/...的相关问题
LCJ

Answers:


123

您的表单除了文件之外不包含任何输入标签,因此在控制器操作中,除了上载的文件(发送到服务器的所有内容)之外,您别无所求。实现此目的的一种方法是包括一个包含模型ID的隐藏标签,该标签可让您从要发布到的控制器操作内的数据存储中检索它(如果用户不应该修改模型,而是使用该模型,只需附加一个文件):

@using (Html.BeginForm("Uploadfile", "Containers", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    @Html.HiddenFor(x => x.Id)
    <input type="file" name="file" id="file" />
    <input type="submit" value="submit" />
}

然后在您的控制器操作中:

[HttpPost]
public ActionResult Uploadfile(int id, HttpPostedFileBase file)
{
    Containers containers = Repository.GetContainers(id);
    ...
}

另一方面,如果要允许用户修改此模型,则需要为要发送到服务器的模型的每个字段包括正确的输入字段:

@using (Html.BeginForm("Uploadfile", "Containers", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    @Html.TextBoxFor(x => x.Prop1)
    @Html.TextBoxFor(x => x.Prop2)
    @Html.TextBoxFor(x => x.Prop3)
    <input type="file" name="file" id="file" />
    <input type="submit" value="submit" />
}

然后您将使用默认的模型装订器从请求中重构此模型:

[HttpPost]
public ActionResult Uploadfile(Container containers, HttpPostedFileBase file)
{
    ...
}

1
我也file等于null并且Request.Files.Count为0,如果同时formAjaxForm和也存在任何差异routeValues
2012年

8

解决了

模型

public class Book
{
public string Title {get;set;}
public string Author {get;set;}
}

控制者

public class BookController : Controller
{
     [HttpPost]
     public ActionResult Create(Book model, IEnumerable<HttpPostedFileBase> fileUpload)
     {
         throw new NotImplementedException();
     }
}

和视图

@using (Html.BeginForm("Create", "Book", FormMethod.Post, new { enctype = "multipart/form-data" }))
{      
     @Html.EditorFor(m => m)

     <input type="file" name="fileUpload[0]" /><br />      
     <input type="file" name="fileUpload[1]" /><br />      
     <input type="file" name="fileUpload[2]" /><br />      

     <input type="submit" name="Submit" id="SubmitMultiply" value="Upload" />
}

注意控制器操作的参数标题必须与输入元素的名称匹配 IEnumerable<HttpPostedFileBase> fileUpload->name="fileUpload[0]"

fileUpload 必须匹配


2
该解决方案是我发现多个文件的唯一解决方案。感谢您分享您的代码。
Rojan Gh。

6

如果您不会总是将图像发布到您的动作中,则可以执行以下操作:

[HttpPost]
public ActionResult Uploadfile(Container container, HttpPostedFileBase file) 
{
    //do container stuff

    if (Request.Files != null)
    {
        foreach (string requestFile in Request.Files)
        {
            HttpPostedFileBase file = Request.Files[requestFile]; 
            if (file.ContentLength > 0)
            {
                string fileName = Path.GetFileName(file.FileName);
                string directory = Server.MapPath("~/App_Data/uploads/");
                if (!Directory.Exists(directory))
                {
                    Directory.CreateDirectory(directory);
                }
                string path = Path.Combine(directory, fileName);
                file.SaveAs(path);
            }
        }
    }

} 

1

对于多个文件;注意输入的较新的“ ”属性:

形成:

@using (Html.BeginForm("FileImport","Import",FormMethod.Post, new {enctype = "multipart/form-data"}))
{
    <label for="files">Filename:</label>
    <input type="file" name="files" multiple="true" id="files" />
    <input type="submit"  />
}

控制器:

[HttpPost]
public ActionResult FileImport(IEnumerable<HttpPostedFileBase> files)
{
    return View();
}

1

第一次从URL下面下载jquery.form.js文件

http://plugins.jquery.com/form/

在cshtml中编写以下代码

@using (Html.BeginForm("Upload", "Home", FormMethod.Post, new { enctype = "multipart/form-data", id = "frmTemplateUpload" }))
{
    <div id="uploadTemplate">

        <input type="text" value="Asif" id="txtname" name="txtName" />


        <div id="dvAddTemplate">
            Add Template
            <br />
            <input type="file" name="file" id="file" tabindex="2" />
            <br />
            <input type="submit" value="Submit" />
            <input type="button" id="btnAttachFileCancel" tabindex="3" value="Cancel" />
        </div>

        <div id="TemplateTree" style="overflow-x: auto;"></div>
    </div>

    <div id="progressBarDiv" style="display: none;">
        <img id="loading-image" src="~/Images/progress-loader.gif" />
    </div>

}


<script type="text/javascript">

    $(document).ready(function () {
        debugger;
        alert('sample');
        var status = $('#status');
        $('#frmTemplateUpload').ajaxForm({
            beforeSend: function () {
                if ($("#file").val() != "") {
                    //$("#uploadTemplate").hide();
                    $("#btnAction").hide();
                    $("#progressBarDiv").show();
                    //progress_run_id = setInterval(progress, 300);
                }
                status.empty();
            },
            success: function () {
                showTemplateManager();
            },
            complete: function (xhr) {
                if ($("#file").val() != "") {
                    var millisecondsToWait = 500;
                    setTimeout(function () {
                        //clearInterval(progress_run_id);
                        $("#uploadTemplate").show();
                        $("#btnAction").show();
                        $("#progressBarDiv").hide();
                    }, millisecondsToWait);
                }
                status.html(xhr.responseText);
            }
        });

    });


</script>

动作方式:-

 public ActionResult Index()
        {
            ViewBag.Message = "Modify this template to jump-start your ASP.NET MVC application.";

            return View();
        }

 public void Upload(HttpPostedFileBase file, string txtname )
        {

            try
            {
                string attachmentFilePath = file.FileName;
                string fileName = attachmentFilePath.Substring(attachmentFilePath.LastIndexOf("\\") + 1);

           }
            catch (Exception ex)
            {

            }
        }
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