问题
I'm wondering if it would be possible at all to upload a file by posting it to a controller action in ASP.NET MVC. The dialog for this upload form will be dynamically generated and will be inside a jQuery dialog in my case.
I'm aware of the file input element but I'm not sure how to send the file to a controller action, not sure how to set the action
parameter
回答1:
Your action should like this:
[HttpPost]
public ActionResult Upload(HttpPostedFileBase file) {
if (file.ContentLength > 0) {
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/App_Data/uploads"), fileName);
file.SaveAs(path);
}
return RedirectToAction("Index");
}
Taken from :http://haacked.com/archive/2010/07/16/uploading-files-with-aspnetmvc.aspx/
Then using jQuery Dialog for file upload:
$dialog.dialog("option", "buttons", {
"Save": function () {
var dlg = $(this);
var formData = new FormData($("#" + formName)[0]);
$.ajax({
url: /Controller/upload,
type: 'POST',
data: formData,
processData: false,
contentType: false,
success: function (response, textStatus, xhr) {
...
}
},
error: function (xhr, status, error) {
....
}
});
},
"Cancel": function () {
$(this).dialog("close");
$(this).empty();
}
});
回答2:
<form id="frmFile" method="post" enctype="multipart/form-data" action="<%=Url.Content("~/WriteSomeServerAction/")%>" >
</form>
//Put this form in modal
and Make Submit event will send file to Action.There you can access file as in Model element as like UploadedFileData .
if (_File.UploadedFileData != null && _File.UploadedFileData.ContentLength > 0)
{
byte[] buffer = new byte[_File.UploadedFileData.ContentLength];
_File.UploadedFileData.InputStream.Read(buffer, 0, buffer.Length);
_File.FileData = System.Text.Encoding.Default.GetString(buffer);
_File.UploadedFileData = null;
}
来源:https://stackoverflow.com/questions/31071374/upload-file-using-jquery-and-post-it-to-controller