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.
Here is my Razor view:
@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:
[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 ?
This question is related to
asp.net-mvc
asp.net-mvc-3
razor
1st download jquery.form.js file from below url
http://plugins.jquery.com/form/
Write below code in cshtml
@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 :-
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)
{
}
}
If you won't always have images posting to your action, you can do something like this:
[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);
}
}
}
}
For multiple files; note the newer "multiple" attribute for input:
Form:
@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:
[HttpPost]
public ActionResult FileImport(IEnumerable<HttpPostedFileBase> files)
{
return View();
}
Solved
Model
public class Book
{
public string Title {get;set;}
public string Author {get;set;}
}
Controller
public class BookController : Controller
{
[HttpPost]
public ActionResult Create(Book model, IEnumerable<HttpPostedFileBase> fileUpload)
{
throw new NotImplementedException();
}
}
And View
@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]"
fileUpload
must match
Source: Stackoverflow.com