asp.net-mvc MVC 3 文件上传和模型绑定
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4784225/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
MVC 3 file upload and model binding
提问by Francis
I have a form upload that works but I would like to pass model information for my database to save the file with a different name of course.
我有一个有效的表单上传，但我想为我的数据库传递模型信息，以使用不同的名称保存文件。
Here is my Razor view:
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" />
}
Here is my Controller:
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");
}
The model information is not passed through to the controller. I have read that I might need to update the model, how would I do this ?
模型信息不会传递到控制器。我已经读到我可能需要更新模型，我该怎么做？
回答by Darin Dimitrov
Your form doesn't contain any input tag other than the file so in your controller action you cannot expect to get anything else than the uploaded file (that's all that's being sent to the server). One way to achieve this would be to include a hidden tag containing the id of the model which will allow you to retrieve it from your datastore inside the controller action you are posting to (use this if the user is not supposed to modify the model but simply attach a file):
您的表单不包含文件以外的任何输入标签，因此在您的控制器操作中，您不能期望获得上传文件以外的任何内容（这就是发送到服务器的所有内容）。实现此目的的一种方法是包含一个包含模型 id 的隐藏标签，这将允许您从您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，则使用它，但是只需附加一个文件）：
@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" />
}
and then in your controller action:
然后在您的控制器操作中：
[HttpPost]
public ActionResult Uploadfile(int id, HttpPostedFileBase file)
{
Containers containers = Repository.GetContainers(id);
...
}
On the other hand if you wanted to allow the user to modify this model then you will need to include the proper input fields for each field of your model that you want to be sent to the server:
另一方面，如果您希望允许用户修改此模型，那么您需要为要发送到服务器的模型的每个字段包含正确的输入字段：
@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" />
}
and then you will have the default model binder reconstruct this model from the request:
然后您将使用默认模型绑定器从请求中重建此模型：
[HttpPost]
public ActionResult Uploadfile(Container containers, HttpPostedFileBase file)
{
...
}
回答by isxaker
Solved
解决了
Model
模型
public class Book
{
public string Title {get;set;}
public string Author {get;set;}
}
Controller
控制器
public class BookController : Controller
{
[HttpPost]
public ActionResult Create(Book model, IEnumerable<HttpPostedFileBase> fileUpload)
{
throw new NotImplementedException();
}
}
And View
并查看
@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" />
}
Note title of parameter from controller action must match with name of input elements
IEnumerable<HttpPostedFileBase> fileUpload-> name="fileUpload[0]"
注意控制器动作的参数标题必须与输入元素的名称匹配
IEnumerable<HttpPostedFileBase> fileUpload->name="fileUpload[0]"
fileUploadmust match
fileUpload必须匹配
回答by jhatcher9999
If you won't always have images posting to your action, you can do something like this:
如果您不会总是将图片发布到您的操作中，您可以执行以下操作：
[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);
}
}
}
}
回答by mcfea
For multiple files; note the newer "multiple" attribute for input:
对于多个文件；注意输入的较新的“ multiple”属性：
Form:
形式：
@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" />
}
Controller:
控制器：
[HttpPost]
public ActionResult FileImport(IEnumerable<HttpPostedFileBase> files)
{
return View();
}
回答by Ace
1st download jquery.form.js file from below url
第一次从下面的 url 下载 jquery.form.js 文件
http://plugins.jquery.com/form/
http://plugins.jquery.com/form/
Write below code in cshtml
用cshtml写下面的代码
@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>
Action method :-
行动方法：-
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)
{
}
}
