Brian Grinstead

<form method="POST" action="http://localhost/" enctype="multipart/form-data">
	File : <input type="file" name="content" size="38" /><br />
	<input type="hidden" name="id" value='fileUpload' />
 </form>

public static class FormUpload
{
	private static readonly Encoding encoding = Encoding.UTF8;
	public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, object> postParameters)
	{
		string formDataBoundary = "-----------------------------28947758029299";
		string contentType = "multipart/form-data; boundary=" + formDataBoundary;
 
		byte[] formData = GetMultipartFormData(postParameters, formDataBoundary);
 
		return PostForm(postUrl, userAgent, contentType, formData);
	}
	private static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, byte[] formData)
	{
		HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;
 
		if (request == null)
		{
			throw new NullReferenceException("request is not a http request");
		}
 
		// Set up the request properties
		request.Method = "POST";
		request.ContentType = contentType;
		request.UserAgent = userAgent;
		request.CookieContainer = new CookieContainer();
		request.ContentLength = formData.Length;  // We need to count how many bytes we're sending. 
 
		using (Stream requestStream = request.GetRequestStream())
		{
			// Push it out there
			requestStream.Write(formData, 0, formData.Length);
			requestStream.Close();
		}
 
		return request.GetResponse() as HttpWebResponse;
	}
 
	private static byte[] GetMultipartFormData(Dictionary<string, object> postParameters, string boundary)
	{
		Stream formDataStream = new System.IO.MemoryStream();
 
		foreach (var param in postParameters)
		{
			if (param.Value is FileParameter)
			{
				FileParameter fileToUpload = (FileParameter)param.Value;
 
				// Add just the first part of this param, since we will write the file data directly to the Stream
				string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\";\r\nContent-Type: {3}\r\n\r\n", 
					boundary, 
					param.Key, 
					fileToUpload.FileName ?? param.Key, 
					fileToUpload.ContentType ?? "application/octet-stream");
 
				formDataStream.Write(encoding.GetBytes(header), 0, header.Length);
 
				// Write the file data directly to the Stream, rather than serializing it to a string.
				formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);
			}
			else
			{
				string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}\r\n", 
					boundary, 
					param.Key, 
					param.Value);
				formDataStream.Write(encoding.GetBytes(postData), 0, postData.Length);
			}
		}
 
		// Add the end of the request
		string footer = "\r\n--" + boundary + "--\r\n";
		formDataStream.Write(encoding.GetBytes(footer), 0, footer.Length);
 
		// Dump the Stream into a byte[]
		formDataStream.Position = 0;
		byte[] formData = new byte[formDataStream.Length];
		formDataStream.Read(formData, 0, formData.Length);
		formDataStream.Close();
 
		return formData;
	}
 
	public class FileParameter
	{
		public byte[] File { get; set; }
		public string FileName { get; set; }
		public string ContentType { get; set; }
		public FileParameter(byte[] file) : this(file, null) { }
		public FileParameter(byte[] file, string filename) : this(file, filename, null) { }
		public FileParameter(byte[] file, string filename, string contenttype) 
		{
			File = file;
			FileName = filename;
			ContentType = contenttype;
		}
	}
}

 
// Read file data
FileStream fs = new FileStream("c:\\people.doc", FileMode.Open, FileAccess.Read);
byte[] data = new byte[fs.Length];
fs.Read(data, 0, data.Length);
fs.Close();
 
// Generate post objects
Dictionary<string, object> postParameters = new Dictionary<string, object>();
postParameters.Add("filename", "People.doc");
postParameters.Add("fileformat", "doc");
postParameters.Add("file", new FormUpload.FileParameter(data, "People.doc", "application/msword"));
 
// Create request and receive response
string postURL = "http://localhost";
string userAgent = "Someone";
HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);
 
// Process response
StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
string fullResponse = responseReader.ReadToEnd();
webResponse.Close();
Response.Write(fullResponse);

39 Responses to “Multipart Form Post in C#”

1. New In Foliotek: Inline File Editing « The Lanit Development Blog Says:
   May 19th, 2009 at 6:35 am

   [...] did a write up about one of the technical challenges that was encountered when adding this feature: generating a multipart form post in C#. Posted in C#, Development, ePortfolio. Tags: Business, Development, foliotek. No Comments [...]

2. Collin Says:
   June 22nd, 2009 at 7:25 am

   This helped me alot, thank you!

3. John Says:
   July 8th, 2009 at 1:35 am

   Thanks for the post! Very helpful.

4. Amrish Shah Says:
   July 9th, 2009 at 11:44 am

   GR8… no words to describe

5. RJ Says:
   July 20th, 2009 at 3:57 pm

   Hi. Great post. I have a quick question about implementing this though. In GetMultipartFormData() -> when processing fileData. Is there a way to have different values for name and filename? right now they are the same value. Also, is there a way to specify the content-type of the file you are uploading? Thanks.

6. Brian Says:
   July 21st, 2009 at 6:57 am

   RJ, good question – that is something that I didn’t need in my specific case (the server I was uploading to ignored the filename and content-type attributes). It would be useful if you were able to specify those attributes, so I decided to add in that feature.

   I updated the code to allow the ability to specify a file name and content-type. Let me know if this works for you!

7. Mark M Says:
   July 21st, 2009 at 12:09 pm

   Great and excellent, also no other words to describe

8. RJ Says:
   July 21st, 2009 at 12:17 pm

   Hey Brian. Thanks for the quick response :) This works. I did have one issue in the method GetMultipartFormData() when you encode the header and post data. I had to create a local variable called encoding:

   Encoding encoding = Encoding.UTF8;

   Which allowed this line to work:

   formDataStream.Write(encoding.GetBytes(header), 0, header.Length);

   I also modified your script to accept a username and password for network credentials in the PostForm() method:

   if (strUserName != null && strPassword != null)
   request.Credentials = new NetworkCredential(strUserName, strPassword);

   I also added a NameValueCollection of possible Cookies as well that get passed into your methods, and in PostForm() I loop through the collection and add them to the CookieContainer:

   NameValueCollection objCookies; //passed into method as a parameter
   request.CookieContainer = new CookieContainer();
   foreach (string strKey in objCookies.AllKeys)
   request.CookieContainer.Add(new Cookie(strKey, objCookies[strKey], “/”, “local.website.com”));

   I needed to send cookies as a way of authenticating to the server (which is a 3rd party community app running on one of our subdomains)

   Thanks!

9. Brian Says:
   July 21st, 2009 at 12:33 pm

   RJ, thanks for the feedback!

   I had forgotten to copy the encoding variable into the code. It is there now – the first line in the class.

   Good idea about adding in the authentication and cookie information. What information are you passing with the cookies? Could you share an example of what you would do to generate objCookies before you call PostForm()?

10. RJ Says:
    July 21st, 2009 at 2:16 pm

    No prob. Like I mentioned earlier, we are using the “form post” to post files to a 3rd party software system running within our domain, so authentication with cookies is valid in our case. Right before I call MultipartFormDataPost(), I create a NameValueCollection of cookies:

    NameValueCollection objCookies = null;
    objCookies = new NameValueCollection();
    objCookies["info] = “uid=johndoe&ts=12999333&apiKey=abcdefg123″;

    FormUpload.MultipartFormDataPost(strUrl, null, null, null, objCookies, objPostParameters);

    Then from here, look at my previous comment on how to set each cookie to the CookieContainer.
    You could, in theory pass in a CookieCollection rather than a NameValueCollection but I just stuck with this approach.

    I believe (in our case) having this cookie set and sent allows us to do simple “shared” domain authentication.
    I’m not sure how this would work when posting to a server in a different domain.

11. Mark M Says:
    July 21st, 2009 at 3:26 pm

    Hello Brian, May I ask if this code only compiled in asp 3.5? I got error with asp.2.0
    complier says ‘FormUpload.FileParameter.File.get’ must declare a body because it is not marked abstract or extern
    ‘FormUpload.FileParameter.File.set’ must declare a body because it is not marked abstract or extern
    Suggestion? Thanks

12. Brian Says:
    July 21st, 2009 at 7:34 pm

    Hi Mark. The code uses the auto-implemented property feature in C# 3.0 and later.

    If you are using an older version, then you should be able to remove them altogether so it would look like:

    public byte[] File;
    public string FileName;
    public string ContentType;

    I think this should work for you. Let me know how it works out.

13. Mark M Says:
    July 22nd, 2009 at 10:43 am

    Thanks Brian, hate to point out again. Below syntax is also available for c# 3.0
    foreach (var param in postParameters)
    {
    if (param.Value is FileParameter)
    {
    FileParameter fileToUpload = (FileParameter)param.Value;
    the var cannot be recognized by the compiler. Will you be able to use another syntax for this purpose?

14. Brian Says:
    July 22nd, 2009 at 11:13 am

    Mark,
    Instead of “var” use:

    foreach (KeyValuePair<string, object> param in postParameters) {  }

    The var is just a convenient shorthand to avoid having to cast the object to that type.

15. Mark M Says:
    July 22nd, 2009 at 12:30 pm

    THanks a lot and I don’t have compile error now. I am trying to upload with following code. But I got the file field empty. Can you see what should be changed?
    // Read file data
    FileStream fs = new FileStream(“E:\\default\\aspnet\\Upload\\Racing.flv”, FileMode.Open, FileAccess.Read);
    byte[] data = new byte[fs.Length];
    fs.Read(data, 0, data.Length);
    fs.Close();

    // Generate post objects
    Dictionary postParameters = new Dictionary();
    postParameters.Add(“filename”, “Racing_001.flv”);
    postParameters.Add(“userlogin”, “tomcattyy”);
    postParameters.Add(“password”, “mark888″);
    postParameters.Add(“title”, “customer 23443″);
    //postParameters.Add(“fileformat”, “doc”);
    postParameters.Add(“file”, new FormUpload.FileParameter(data, “E:\\default\\aspnet\\Upload\\Racing.flv”, “flv-application/octet-stream”));

    // Create request and receive response
    string postURL = “http://uploads.blip.tv/file/post”;
    string userAgent = “Someone”;
    HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);

    // Process response
    StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
    string fullResponse = responseReader.ReadToEnd();
    webResponse.Close();
    Response.Write(fullResponse);

16. Brian Says:
    July 22nd, 2009 at 12:40 pm

    Hard to say. Maybe the Filename is causing problems with the post since is has those “\\” in it? Try it with just “Racing.flv”.

    What do you mean by “got the file field empty”? Did it work before? Does the rest of the post data go through to your postURL?

17. Mark M Says:
    July 22nd, 2009 at 1:56 pm

    Yes, it takes the user / password / title but file name field is blank, .
    I have changed the file name in
    postParameters.Add(“file”, new FormUpload.FileParameter(data, “Racing.flv”, “flv-application/octet-stream”));
    please take a look at http://208.75.252.245/video_blipup.aspx

    Thanks

    Mark

18. Brian Says:
    July 22nd, 2009 at 2:28 pm

    Are you doing uploads from the user’s computer onto your server, or are you uploading a file from your server to another server?

    If you are doing uploads from the users’s computer, check out How to upload a file to a Web server in ASP.NET by using Visual C# .NET.

    If you are trying to move files from your server to another server, is it only the **file name** that is not coming through, or is the **entire file** not being uploaded?

19. Mark M Says:
    July 22nd, 2009 at 2:44 pm

    I am moving video files from my server to another server. the entire file did not went through.
    From the last post link, the file field is blank and I log into blip.tv confirm those files not been upload either. Thanks

20. Brian Says:
    July 22nd, 2009 at 3:23 pm

    Mark, the interface on that page is a little confusing to me. I’m not sure if I am supposed to be uploading a new video, or how to make it upload the one already on the server.

    If you have access to the destination server, maybe you could check what is being received and print out some output to diagnose the problem.
    If you don’t have access to the destination server, maybe you could point it to a script you do have access to (change the postURL) and check the Request object to see what is coming through. Specifically, check the Request.Files object to see what is going on with it.

    … or maybe it is expecting the parameter name to be something other than “file”?

    Let me know what you figure out!

21. Mark M Says:
    July 22nd, 2009 at 3:54 pm

    thanks Brian, please go ahead uploading whatever video to test. I will change the postURL to test.
    blip.tv API is located here
    http://wiki.blip.tv/index.php/REST_Upload_API
    if you want to take a look. they only have a python sample there

22. JP Toto Says:
    July 28th, 2009 at 2:30 pm

    This was fantastically helpful, Brian. Thanks so much for publishing this. I’m still studying to see if I can learn the important parts of a multi-part form post. Cheers!

23. JP Toto Says:
    August 15th, 2009 at 11:16 am

    Brian,

    I took the example you posted here and started a GitHub project based on it where I’m adding some more features and refactoring the code a bit. http://github.com/jptoto/multipart_form_poster/tree/master I give full credit in the README file and a link to your blog post. I hope this ok.

    This post REALLY helped me get moving on a project I’m working on. It was a big help for POSTing to an external API. Thanks again.

    Cheers!

24. Joe G Says:
    September 24th, 2009 at 10:43 am

    Brian,

    In order to work with multiple file uploads, I had to add a CR+LF to the end of each FileParameter stream. In the code, I added the line:

    formDataStream.Write(encoding.GetBytes("\r\n"), 0, 2);

    to GetMultipartFormData, after the line:

    formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);

    Now, with two FileParameters in the Dictionary, I see both files in the Request.Files collection on the other side.

    Cheers,
    -Joe

25. SlideShare .NET API Wrapper | Frederik Vig - ASP.NET developer Says:
    October 18th, 2009 at 6:32 am

    [...] exposed and made available for you to use. Big thanks to Brian Grinstead for his blog post Multipart Form Post in C# and for Gaurav Gupta for creating the wrapper class for the first version, which this code is based [...]

26. Mike Ferreira Says:
    November 5th, 2009 at 8:56 am

    Excellent example. Needed a VB.Net solution, so I converted to VB… Here is the code is interested.

    Imports System.Text
    Imports System.Net
    Imports System.IO
     
    Public Class FormUpload
     
        Private ReadOnly encoding As Encoding = encoding.UTF8
        Public Function MultipartFormDataPost(ByVal postUrl As String, _
                                              ByVal userAgent As String, _
                                              ByVal postParameters As Dictionary(Of String, Object)) _
                                              As HttpWebResponse
            Dim formDataBoundary As String = "-----------------------------19330813700727"
            Dim contentType As String = "multipart/form-data; boundary=" + formDataBoundary
     
            Dim formData As Byte() = GetMultipartFormData(postParameters, formDataBoundary)
     
            Return PostForm(postUrl, userAgent, contentType, formData)
     
        End Function
     
        Private Function PostForm(ByVal postUrl As String, _
                                  ByVal userAgent As String, _
                                  ByVal contentType As String, _
                                  ByVal formData As Byte()) _
                                  As HttpWebResponse
     
            Try
                Dim request As HttpWebRequest = WebRequest.Create(postUrl)
     
                If request Is Nothing Then
                    Throw New NullReferenceException("request is not a http request")
                End If
     
                'Set up the request properties
                request.Method = "POST"
                request.ContentType = contentType
                request.UserAgent = userAgent
                request.CookieContainer = New CookieContainer()
                request.ContentLength = formData.Length  'We need to count how many bytes we're sending. 
     
                Using requestStream As Stream = request.GetRequestStream()
                    'Push it out there
                    requestStream.Write(formData, 0, formData.Length)
                    requestStream.Close()
                End Using
     
                Return request.GetResponse
            Catch ex As Exception
                MsgBox("Exception: " & ex.Message & vbCrLf & "Stack: " & ex.StackTrace)
                Throw ex
                Return Nothing
            End Try
     
        End Function
     
     
        Private Function GetMultipartFormData(ByVal postParameters As Dictionary(Of String, Object), _
                                              ByVal boundary As String) _
                                              As Byte()
     
            Dim formDataStream As Stream = New System.IO.MemoryStream()
            For Each param As KeyValuePair(Of String, Object) In postParameters
     
                If TypeOf (param.Value) Is FileParameter Then
                    Dim fileToUpload As FileParameter = param.Value
                    'Add just the first part of this param, since we will write the file data directly to the Stream
                    Dim header As String = "--" & boundary & vbCrLf & _
                                           "Content-Disposition: form-data; name=""" & param.Key & """; filename=""" & fileToUpload.FileName & """;" & vbCrLf & _
                                           "Content-Type: " & fileToUpload.ContentType & vbCrLf & _
                                           "Content-Transfer-Encoding: binary" & vbCrLf & _
                                           vbCrLf
     
                    formDataStream.Write(encoding.GetBytes(header), 0, header.Length)
     
                    'Write the file data directly to the Stream, rather than serializing it to a string.
                    formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length)
     
                Else
                    Dim postData As String = "--" & boundary & vbCrLf & _
                                             "Content-Disposition: form-data; name=""" & param.Key & """" & vbCrLf & _
                                             "Content-Type: text/plain; charset=windows-1252" & vbCrLf & _
                                             "Content-Transfer-Encoding: 8bit" & vbCrLf & _
                                             vbCrLf & _
                                             param.Value & vbCrLf
                    formDataStream.Write(encoding.GetBytes(postData), 0, postData.Length)
     
                End If
     
            Next
     
            'Add the end of the request
            Dim footer As String = vbCrLf & "--" & boundary & "--" & vbCrLf
            formDataStream.Write(encoding.GetBytes(footer), 0, footer.Length)
     
            'Dump the Stream into a byte()
            formDataStream.Position = 0
            Dim formData As Byte() = New Byte(formDataStream.Length) {}
            formDataStream.Read(formData, 0, formData.Length)
            formDataStream.Close()
     
            Return formData
     
        End Function
     
     
        Public Class FileParameter
            Private pFile As Byte()
            Private pFileName As String
            Private pContentType As String
     
            Public Property File() As Byte()
                Get
                    Return pFile
                End Get
                Set(ByVal value As Byte())
                    pFile = value
                End Set
            End Property
     
            Public Property FileName() As String
                Get
                    Return pFileName
                End Get
                Set(ByVal value As String)
                    pFileName = value
                End Set
            End Property
     
            Public Property ContentType() As String
                Get
                    Return pContentType
                End Get
                Set(ByVal value As String)
                    pContentType = value
                End Set
            End Property
     
            Sub New(ByVal file As Byte())
                Me.New(file, Nothing)
            End Sub
            Sub New(ByVal file As Byte(), ByVal filename As String)
                Me.New(file, filename, Nothing)
            End Sub
            Sub New(ByVal file As Byte(), ByVal filename As String, ByVal contenttype As String)
                Me.File = file
                Me.FileName = filename
                Me.ContentType = contenttype
            End Sub
     
        End Class
     
    End Class

27. lee Says:
    November 8th, 2009 at 1:21 pm

    Newbie question … would this code also work on a Windows Mobile 6 device ?
    thanks

28. Brian Says:
    November 9th, 2009 at 7:25 am

    Mike,
    Thanks for taking the time to post that code after you converted it over to VB.Net. Hopefully it is working for you in your project.

29. Brian Says:
    November 9th, 2009 at 7:32 am

    Lee,
    As far as I know, it should work on Windows Mobile 6. The classes that it uses are mostly in System.Net and System.IO. I’ve never done any Windows Mobile development, so I couldn’t tell you for sure either way.

    Take a look at creating your first Windows Mobile 6 Application haven’t gotten the SDK set up yet. Let me know if you try it, and if you needed to make any changes for it to work.

30. Upload a File via WebRequest Using CSharp Says:
    November 10th, 2009 at 7:15 am

    [...] Brian Grinstead » Blog Archive » Multipart Form Post in C# – So, I needed to roll my own form post. Here is the Multipart Form RFC and the W3C Specification for multipart/form data. After reading these links and searching some forums, here is what I came up with. ….. userAgent, contentType, formData) End Function Private Function PostForm(ByVal postUrl As String, _ ByVal userAgent As String, _ ByVal contentType As String, _ ByVal formData As Byte()) _ As HttpWebResponse Try Dim request As HttpWebRequest = WebRequest. … [...]

31. John Sheehan Says:
    December 1st, 2009 at 6:10 pm

    Hi Brian, can you email me in regards to the code in this post? Or can you acknowledge if this code is under any sort of license?

32. Pkpo Says:
    December 14th, 2009 at 9:16 am

    I love you !

33. Corey Says:
    February 9th, 2010 at 7:21 am

    This class worked great for posting a multipart form in c sharp. Thanks!

34. Mithun Says:
    February 10th, 2010 at 5:24 am

    Awsome post!!! Thanks for helping us out of the drudgery…. This saved a lot of time in my work…. May be this should be added as a part of the C# library.

35. Turkel Says:
    March 6th, 2010 at 1:57 am

    Hi Mr Brian

    I need your help about that post.

    First

    I have php file that gets data and file from HTML form and uploads file to specific directory by $_FILES methods..

    Second

    I have windos aplication (c#).User could drag and drop any file to listview.When he/She clicks send button aplication sends datas of user to php file By POST method and it (php file) upadtes the database.

    My problem is I wont to send file in listview drag and dropeed by user to php file as It could get file by $_FILES method and acces it.Sory about my english

    thanks.what do I have to do by your post?

36. Matt Says:
    April 8th, 2010 at 6:09 pm

    Brian – outstanding! Really helped…

    Matt

37. Pykaso Says:
    May 10th, 2010 at 6:29 am

    For UTF-8 encoded form data im using:

    formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));

    instead of:

    formDataStream.Write(encoding.GetBytes(postData), 0, postData.Length);

38. diss3ntive Says:
    June 22nd, 2010 at 7:25 pm

    I…..Love……Youuuuuu

39. Stephen Says:
    August 31st, 2010 at 10:57 am

    Absolutely fantastic post. Hve been working on this for the past two days to no avail. I’ve been trying to programatically post files to a unix server and it does not help when when the Java guys
    a) Decide to use a form post method for posting files when FTP would adequately do the Job
    b) Give you Zero support when their unix server just returns “403″ bad request, apparently they get nothing in the logs

    If anyone is interested in using basic authentication with this code,

    after the line

    request.ContentLength = formData.Length;

    like I had to add the following to get it to work for me,

    request.PreAuthenticate = true;
    request.AuthenticationLevel = System.Net.Security.AuthenticationLevel.MutualAuthRequested;
    request.Headers.Add("Authorization", "Basic " + Convert.ToBase64String(System.Text.Encoding.Default.GetBytes("username" + ":" + "password")));

    I also wrapped the FileStream and StreamReader object in my caller method within using blocks, which is just decorative/programming style stuff really if you wish.

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