Я пытался использовать приведенный здесь пример Как сделать сообщение формы ASP.NET MVC Ajax с данными multipart/form?
Но я продолжаю получать сообщение об ошибке "fail"
Index.cshtml
<script src="~/Scripts/jquery-1.8.2.min.js"></script>
<script src="~/Scripts/jquery.unobtrusive-ajax.min.js"></script>
<h2>Files Upload</h2>
<script type="text/javascript">
$(function() {
$("#form0").submit(function(event) {
var dataString;
event.preventDefault();
var action = $("#form0").attr("action");
if ($("#form0").attr("enctype") == "multipart/form-data") {
//this only works in some browsers.
//purpose? to submit files over ajax. because screw iframes.
//also, we need to call .get(0) on the jQuery element to turn it into a regular DOM element so that FormData can use it.
dataString = new FormData($("#form0").get(0));
contentType = false;
processData = false;
} else {
// regular form, do your own thing if you need it
}
$.ajax({
type: "POST",
url: action,
data: dataString,
dataType: "json", //change to your own, else read my note above on enabling the JsonValueProviderFactory in MVC
contentType: contentType,
processData: processData,
success: function(data) {
//BTW, data is one of the worst names you can make for a variable
},
error: function(jqXHR, textStatus, errorThrown) {
//do your own thing
alert("fail");
}
});
}); //end .submit()
});
</script>
<div id="uploadDiv">
@Html.Action("Files", "Home")
</div>
@using (Ajax.BeginForm("Files", "Home", new AjaxOptions { UpdateTargetId = "uploadDiv", HttpMethod = "Post" }, new { enctype = "multipart/form-data", @id="form0"}))
{
<div>
<div>Upload new file:
<input type="file" name="file" /></div>
<input type="submit" value="Save" />
</div>
}
<br />
контроллер
public PartialViewResult Files(HttpPostedFileBase file)
{
IEnumerable<string> files;
if ((file != null) && (file.ContentLength > 0))
{
string fileName = file.FileName;
string saveLocation = @"D:\Files";
string fullFilePath = Path.Combine(saveLocation, fileName);
try
{
file.SaveAs(fullFilePath);
FileInfo fileInfo = new FileInfo(fullFilePath);
file.InputStream.Read(new byte[fileInfo.Length], 0, file.ContentLength);
}
catch (Exception e)
{
TempData["FileUpload"] = e.Message;
return PartialView();
}
files = Directory.GetFiles(@"D:\Files\");
return PartialView(files);
}
else
{
files = Directory.GetFiles(@"D:\Files\");
return PartialView(files);
}
}
Files.cshtml
@model IEnumerable<string>
@foreach (string f in Model)
{
<p>@f</p>
}
Global.asax
ValueProviderFactories.Factories.Add(new JsonValueProviderFactory());