Problem
(Note: this question pertains to ASP.NET MVC 3.0, which was launched in 2011, not ASP.NET Core 3.0, which was released in 2019.)
In asp.net mvc, I want to upload a file. How do I use the html input file control to upload the file?
Asked by user637197
Solution #1
A file input control isn’t used. In ASP.NET MVC, server-side controls are not used. Check out the following blog post for an example of how to do it in ASP.NET MVC.
As a result, you’d start by constructing an HTML form with a file input:
@using (Html.BeginForm("Index", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<input type="file" name="file" />
<input type="submit" value="OK" />
}
After that, you’d need a controller to manage the upload:
public class HomeController : Controller
{
// This action renders the form
public ActionResult Index()
{
return View();
}
// This action handles the form POST and the upload
[HttpPost]
public ActionResult Index(HttpPostedFileBase file)
{
// Verify that the user selected a file
if (file != null && file.ContentLength > 0)
{
// extract only the filename
var fileName = Path.GetFileName(file.FileName);
// store the file inside ~/App_Data/uploads folder
var path = Path.Combine(Server.MapPath("~/App_Data/uploads"), fileName);
file.SaveAs(path);
}
// redirect back to the index action to show the form once again
return RedirectToAction("Index");
}
}
Answered by Darin Dimitrov
Solution #2
to convert to byte[] (for example, to save to a database):
using (MemoryStream ms = new MemoryStream()) {
file.InputStream.CopyTo(ms);
byte[] array = ms.GetBuffer();
}
You can use this class, which was adapted from here, to transfer the input stream directly into the database without saving it in memory:
public class VarbinaryStream : Stream {
private SqlConnection _Connection;
private string _TableName;
private string _BinaryColumn;
private string _KeyColumn;
private int _KeyValue;
private long _Offset;
private SqlDataReader _SQLReader;
private long _SQLReadPosition;
private bool _AllowedToRead = false;
public VarbinaryStream(
string ConnectionString,
string TableName,
string BinaryColumn,
string KeyColumn,
int KeyValue,
bool AllowRead = false)
{
// create own connection with the connection string.
_Connection = new SqlConnection(ConnectionString);
_TableName = TableName;
_BinaryColumn = BinaryColumn;
_KeyColumn = KeyColumn;
_KeyValue = KeyValue;
// only query the database for a result if we are going to be reading, otherwise skip.
_AllowedToRead = AllowRead;
if (_AllowedToRead == true)
{
try
{
if (_Connection.State != ConnectionState.Open)
_Connection.Open();
SqlCommand cmd = new SqlCommand(
@"SELECT TOP 1 [" + _BinaryColumn + @"]
FROM [dbo].[" + _TableName + @"]
WHERE [" + _KeyColumn + "] = @id",
_Connection);
cmd.Parameters.Add(new SqlParameter("@id", _KeyValue));
_SQLReader = cmd.ExecuteReader(
CommandBehavior.SequentialAccess |
CommandBehavior.SingleResult |
CommandBehavior.SingleRow |
CommandBehavior.CloseConnection);
_SQLReader.Read();
}
catch (Exception e)
{
// log errors here
}
}
}
// this method will be called as part of the Stream ímplementation when we try to write to our VarbinaryStream class.
public override void Write(byte[] buffer, int index, int count)
{
try
{
if (_Connection.State != ConnectionState.Open)
_Connection.Open();
if (_Offset == 0)
{
// for the first write we just send the bytes to the Column
SqlCommand cmd = new SqlCommand(
@"UPDATE [dbo].[" + _TableName + @"]
SET [" + _BinaryColumn + @"] = @firstchunk
WHERE [" + _KeyColumn + "] = @id",
_Connection);
cmd.Parameters.Add(new SqlParameter("@firstchunk", buffer));
cmd.Parameters.Add(new SqlParameter("@id", _KeyValue));
cmd.ExecuteNonQuery();
_Offset = count;
}
else
{
// for all updates after the first one we use the TSQL command .WRITE() to append the data in the database
SqlCommand cmd = new SqlCommand(
@"UPDATE [dbo].[" + _TableName + @"]
SET [" + _BinaryColumn + @"].WRITE(@chunk, NULL, @length)
WHERE [" + _KeyColumn + "] = @id",
_Connection);
cmd.Parameters.Add(new SqlParameter("@chunk", buffer));
cmd.Parameters.Add(new SqlParameter("@length", count));
cmd.Parameters.Add(new SqlParameter("@id", _KeyValue));
cmd.ExecuteNonQuery();
_Offset += count;
}
}
catch (Exception e)
{
// log errors here
}
}
// this method will be called as part of the Stream ímplementation when we try to read from our VarbinaryStream class.
public override int Read(byte[] buffer, int offset, int count)
{
try
{
long bytesRead = _SQLReader.GetBytes(0, _SQLReadPosition, buffer, offset, count);
_SQLReadPosition += bytesRead;
return (int)bytesRead;
}
catch (Exception e)
{
// log errors here
}
return -1;
}
public override bool CanRead
{
get { return _AllowedToRead; }
}
protected override void Dispose(bool disposing)
{
if (_Connection != null)
{
if (_Connection.State != ConnectionState.Closed)
try { _Connection.Close(); }
catch { }
_Connection.Dispose();
}
base.Dispose(disposing);
}
#region unimplemented methods
public override bool CanSeek
{
get { return false; }
}
public override bool CanWrite
{
get { return true; }
}
public override void Flush()
{
throw new NotImplementedException();
}
public override long Length
{
get { throw new NotImplementedException(); }
}
public override long Position
{
get
{
throw new NotImplementedException();
}
set
{
throw new NotImplementedException();
}
}
public override long Seek(long offset, SeekOrigin origin)
{
throw new NotImplementedException();
}
public override void SetLength(long value)
{
throw new NotImplementedException();
}
#endregion unimplemented methods }
and the usage:
using (var filestream = new VarbinaryStream(
"Connection_String",
"Table_Name",
"Varbinary_Column_name",
"Key_Column_Name",
keyValueId,
true))
{
postedFile.InputStream.CopyTo(filestream);
}
Answered by Arthur
Solution #3
A different way to convert to byte[] (for saving to DB).
@Arthur’s approach is fairly effective, but it does not copy exactly, therefore MS Office documents may fail to open after being retrieved from the database. MemoryStream.GetBuffer() may produce extra empty bytes at the end of the byte[], however this can be avoided by instead using MemoryStream.ToArray(). This solution, on the other hand, worked perfectly for all file types:
using (var binaryReader = new BinaryReader(file.InputStream))
{
byte[] array = binaryReader.ReadBytes(file.ContentLength);
}
Here’s the complete code:
Document Class:
public class Document
{
public int? DocumentID { get; set; }
public string FileName { get; set; }
public byte[] Data { get; set; }
public string ContentType { get; set; }
public int? ContentLength { get; set; }
public Document()
{
DocumentID = 0;
FileName = "New File";
Data = new byte[] { };
ContentType = "";
ContentLength = 0;
}
}
File Download:
[HttpGet]
public ActionResult GetDocument(int? documentID)
{
// Get document from database
var doc = dataLayer.GetDocument(documentID);
// Convert to ContentDisposition
var cd = new System.Net.Mime.ContentDisposition
{
FileName = doc.FileName,
// Prompt the user for downloading; set to true if you want
// the browser to try to show the file 'inline' (display in-browser
// without prompting to download file). Set to false if you
// want to always prompt them to download the file.
Inline = true,
};
Response.AppendHeader("Content-Disposition", cd.ToString());
// View document
return File(doc.Data, doc.ContentType);
}
File Upload:
[HttpPost]
public ActionResult GetDocument(HttpPostedFileBase file)
{
// Verify that the user selected a file
if (file != null && file.ContentLength > 0)
{
// Get file info
var fileName = Path.GetFileName(file.FileName);
var contentLength = file.ContentLength;
var contentType = file.ContentType;
// Get file data
byte[] data = new byte[] { };
using (var binaryReader = new BinaryReader(file.InputStream))
{
data = binaryReader.ReadBytes(file.ContentLength);
}
// Save to database
Document doc = new Document()
{
FileName = fileName,
Data = data,
ContentType = contentType,
ContentLength = contentLength,
};
dataLayer.SaveDocument(doc);
// Show success ...
return RedirectToAction("Index");
}
else
{
// Show error ...
return View("Foo");
}
}
View (snippet):
@using (Html.BeginForm("GetDocument", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<input type="file" name="file" />
<input type="submit" value="Upload File" />
}
Answered by Lane
Solution #4
Often, you’ll want to pass a viewmodel in addition to the main file. Other important features can be found in the code below:
The following code might be used to accomplish this:
[HttpPost]
public ActionResult Index(MyViewModel viewModel)
{
// if file's content length is zero or no files submitted
if (Request.Files.Count != 1 || Request.Files[0].ContentLength == 0)
{
ModelState.AddModelError("uploadError", "File's length is zero, or no files found");
return View(viewModel);
}
// check the file size (max 4 Mb)
if (Request.Files[0].ContentLength > 1024 * 1024 * 4)
{
ModelState.AddModelError("uploadError", "File size can't exceed 4 MB");
return View(viewModel);
}
// check the file size (min 100 bytes)
if (Request.Files[0].ContentLength < 100)
{
ModelState.AddModelError("uploadError", "File size is too small");
return View(viewModel);
}
// check file extension
string extension = Path.GetExtension(Request.Files[0].FileName).ToLower();
if (extension != ".pdf" && extension != ".doc" && extension != ".docx" && extension != ".rtf" && extension != ".txt")
{
ModelState.AddModelError("uploadError", "Supported file extensions: pdf, doc, docx, rtf, txt");
return View(viewModel);
}
// extract only the filename
var fileName = Path.GetFileName(Request.Files[0].FileName);
// store the file inside ~/App_Data/uploads folder
var path = Path.Combine(Server.MapPath("~/App_Data/uploads"), fileName);
try
{
if (System.IO.File.Exists(path))
System.IO.File.Delete(path);
Request.Files[0].SaveAs(path);
}
catch (Exception)
{
ModelState.AddModelError("uploadError", "Can't save file to disk");
}
if(ModelState.IsValid)
{
// put your logic here
return View("Success");
}
return View(viewModel);
}
Check to see whether you have
@Html.ValidationMessage("uploadError")
Validation problems can be found in your view.
Keep in mind that the default maximum request length is 4MB (maxRequestLength = 4096), thus you’ll need to alter this option in web.config to upload larger files:
<system.web>
<httpRuntime maxRequestLength="40960" executionTimeout="1100" />
(In this case, 40960 = 40 MB).
The execution timeout is defined as the total number of seconds. You might wish to alter it so that large files can be uploaded.
Answered by Roman Pushkin
Solution #5
In the view:
<form action="Categories/Upload" enctype="multipart/form-data" method="post">
<input type="file" name="Image">
<input type="submit" value="Save">
</form>
when in the controller, the following code:
public ActionResult Upload()
{
foreach (string file in Request.Files)
{
var hpf = this.Request.Files[file];
if (hpf.ContentLength == 0)
{
continue;
}
string savedFileName = Path.Combine(
AppDomain.CurrentDomain.BaseDirectory, "PutYourUploadDirectoryHere");
savedFileName = Path.Combine(savedFileName, Path.GetFileName(hpf.FileName));
hpf.SaveAs(savedFileName);
}
...
}
Answered by Muhammad Soliman
Post is based on https://stackoverflow.com/questions/5193842/file-upload-asp-net-mvc-3-0