Multipart Form Post in C#

I recently had to access a web API through C Sharp that required a file upload. This is pretty easy if you have an HTML page with a form tag and you want a user to directly upload the file.

<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>

However, this is not always a reasonable path to take. Sometimes you may be wanting to access a file that is already in a system and you don’t want a new upload. If you are accessing an external API, this is probably always the case. Unfortunately, building this post using C# is not quite as straightforward. I first tried using the WebClient UploadFile method, but it didn’t fit my needs because I wanted to upload form values (id, filename, other API specific parameters) in addition to just a file.

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.

Update: This post has gotten a great response from all the readers who have taken the time to comment and contribute. I would like to take this opportunity to promote the best REST Client for .NET, RestSharp. John Sheehan has implemented this technique using code from this post, which can be seen on github (just look for WriteMultipartFormData). He has also done a great job implementing other basic REST operations in a fully tested suite. I would recommend reading the rest of the post to figure out what is going on behind the scenes, but you might consider using RestSharp in a production environment. Thanks for reading!

Note: If anyone is interested in this code in Visual Basic, reader Mike Ferreira converted the code into VB.Net in a comment below.

// Implements multipart/form-data POST in C# http://www.ietf.org/rfc/rfc2388.txt
// http://www.briangrinstead.com/blog/multipart-form-post-in-c
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 = String.Format("----------{0:N}", Guid.NewGuid());
        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;
 
        // You could add authentication here as well if needed:
        // request.PreAuthenticate = true;
        // request.AuthenticationLevel = System.Net.Security.AuthenticationLevel.MutualAuthRequested;
        // request.Headers.Add("Authorization", "Basic " + Convert.ToBase64String(System.Text.Encoding.Default.GetBytes("username" + ":" + "password")));
 
        // Send the form data to the request.
        using (Stream requestStream = request.GetRequestStream())
        {
            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();
        bool needsCLRF = false;
 
        foreach (var param in postParameters)
        {
            // Thanks to feedback from commenters, add a CRLF to allow multiple parameters to be added.
            // Skip it on the first parameter, add it to subsequent parameters.
            if (needsCLRF)
                formDataStream.Write(encoding.GetBytes("\r\n"), 0, encoding.GetByteCount("\r\n"));
 
            needsCLRF = true;
 
            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, encoding.GetByteCount(header));
 
                // 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}",
                    boundary,
                    param.Key,
                    param.Value);
                formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));
            }
        }
 
        // Add the end of the request.  Start with a newline
        string footer = "\r\n--" + boundary + "--\r\n";
        formDataStream.Write(encoding.GetBytes(footer), 0, encoding.GetByteCount(footer));
 
        // 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;
        }
    }
}

Here is the code to call the MultipartFormDataPost function with multiple parameters, including a file.

 
// 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);

Hopefully this code can help someone, figuring out exactly where to place the boundary and newlines in between form key-value pairs caused a little bit of grief during development. This is some functionality that would be really nice inside of the language library, but it seems like in most languages this is something you end up coding yourself.

Tags: asp.net, csharp, programming, vb.net

This entry was posted on Friday, May 8th, 2009 at 7:58 am and is filed under All, Web Development. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

78 Responses to “Multipart Form Post in C#”

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 [...]
Collin Says:
June 22nd, 2009 at 7:25 am
This helped me alot, thank you!
John Says:
July 8th, 2009 at 1:35 am
Thanks for the post! Very helpful.
Amrish Shah Says:
July 9th, 2009 at 11:44 am
GR8… no words to describe
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.
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!
Mark M Says:
July 21st, 2009 at 12:09 pm
Great and excellent, also no other words to describe
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!
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()?
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.
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
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.
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?
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.
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);
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?
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
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?
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
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!
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
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!
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!
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
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 [...]

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: " &amp; ex.Message &amp; vbCrLf &amp; "Stack: " &amp; 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 = "--" &amp; boundary &amp; vbCrLf &amp; _
                                       "Content-Disposition: form-data; name=""" &amp; param.Key &amp; """; filename=""" &amp; fileToUpload.FileName &amp; """;" &amp; vbCrLf &amp; _
                                       "Content-Type: " &amp; fileToUpload.ContentType &amp; vbCrLf &amp; _
                                       "Content-Transfer-Encoding: binary" &amp; vbCrLf &amp; _
                                       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 = "--" &amp; boundary &amp; vbCrLf &amp; _
                                         "Content-Disposition: form-data; name=""" &amp; param.Key &amp; """" &amp; vbCrLf &amp; _
                                         "Content-Type: text/plain; charset=windows-1252" &amp; vbCrLf &amp; _
                                         "Content-Transfer-Encoding: 8bit" &amp; vbCrLf &amp; _
                                         vbCrLf &amp; _
                                         param.Value &amp; vbCrLf
                formDataStream.Write(encoding.GetBytes(postData), 0, postData.Length)
 
            End If
 
        Next
 
        'Add the end of the request
        Dim footer As String = vbCrLf &amp; "--" &amp; boundary &amp; "--" &amp; 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

lee Says:
November 8th, 2009 at 1:21 pm
Newbie question … would this code also work on a Windows Mobile 6 device ?
thanks
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.
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.
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. … [...]
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?
Pkpo Says:
December 14th, 2009 at 9:16 am
I love you !
Corey Says:
February 9th, 2010 at 7:21 am
This class worked great for posting a multipart form in c sharp. Thanks!
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.
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?
Matt Says:
April 8th, 2010 at 6:09 pm
Brian – outstanding! Really helped…

Matt

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);

diss3ntive Says:
June 22nd, 2010 at 7:25 pm
I…..Love……Youuuuuu
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.

Pykaso Says:
September 30th, 2010 at 5:15 am

When sending multiple files with this class, last file have extra 2 bytes “\r\n” at end of the file.

Test for last 2 bytes before appending footer ..

byte[] lastB = new byte[2];
formDataStream.Position = formDataStream.Length - 2;
formDataStream.Read(lastB, 0, 2);
 
string footer = "";
if (lastB[0] != 13 || lastB[1] != 10)
{
   footer += "\r\n";
}
// Add the end of the request
footer = "--" + boundary + "--\r\n";

Derek Says:
December 16th, 2010 at 4:31 pm
Hi Brian,

Thank you for this helpful and informative post. There was one minor problem I noticed: Shouldn’t a CRLF be appended after the filedata is written in GetMultipartFormData()? According to the RFC specification “the encapsulation boundary must occur at the beginning of a line.” Sorry to be picky, but I thought it would be help for other people reusing this code ;)
Brian Says:
December 16th, 2010 at 4:58 pm
Derek,
Thanks for the feedback. I have seen that in earlier comments, but for some reason yours finally made me take another look at it and see where the mistake is. Is this the fix you had in mind?
```
formDataStream.Write(encoding.GetBytes("\r\n"), 0, 2);
```
I would also like to point out that RestSharp is an excellent library that uses the functionality from this post. You can see that in action here: https://github.com/johnsheehan/RestSharp/blob/master/RestSharp/Http.Sync.cs#L162. If you need a robust REST Client, it is a great choice.

Thanks!
Bogdan Says:
February 1st, 2011 at 10:07 am
Thanks a lot for this! Had to do something similar for a client and you’ve helped a lot!
Marek Says:
February 2nd, 2011 at 4:28 pm
Brian,

Thanks very much. i tried to build mine from scratch building the headers, I was close, but not close enough! Your code worked pretty well straight off the bat, just plugged in my values and off I went. Spent a whole day on this before finding yours. Excellent Post!

THANKS AGAIN!

Marek
Mochaz Says:
February 9th, 2011 at 5:22 am
I have tried the following code, but I do not get test.jpg in wwwroot folder
What’s wrong with my code?
Do I need to put html file or other in the wwwroot folder?

Thanks in advance
:

// Read file data
FileStream fs = new FileStream(“c:\\test.jpg”, 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”, “test.txt”);
postParameters.Add(“fileformat”, “txt”);
postParameters.Add(“file”, new FormUpload.FileParameter(data, “test.jpg”, “image/jpeg”));

// Create request and receive response
string postURL = “http://localhost”;
string userAgent = “Someone”;
HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);
Brian Says:
February 9th, 2011 at 12:06 pm
Mochaz,
That code will post your test.jpg file to http://localhost. You need to have your script at http://localhost handle this file in some way (like saving it in the wwwroot folder or something else. A script running in http://localhost/Default.aspx should be able to access the Request.Files collection and save it.

For more information on this, see listing 6: “uploading multiple files to the server” on MSDN here: http://msdn.microsoft.com/en-us/library/aa479405.aspx
Deathtospam Says:
February 22nd, 2011 at 2:56 pm
@Pykaso: Why do you call the encoding.GetByteCount(postData) method instead of using the postData.Length property, when dealing with UTF8 encoding?
Russell Says:
April 13th, 2011 at 7:00 am
Brian,

I’m new to C# and was looking for a solution for sending data/image to a remote PHP script. Your code was very helpful.

Thank you!
Multipart Form Posts in .NET | objetoa.net Says:
May 23rd, 2011 at 9:22 pm
[...] I added some arbitrary number of POST parameters.After awhile I found Brian Grinstead‘s post about how he had constructed his own class that enabled users to create their own custom-formed [...]
fabiim Says:
May 27th, 2011 at 2:28 pm
How can i adapt this to wait for the entire file to get trough ?
My request is timing out . Altough i see the server answer in fidler after the timeout exception being thrown .
belle Says:
May 31st, 2011 at 6:42 pm
Hi Brian,
Would this work for uploads from a client machine to a remote server? I’ve been breaking my head trying to figure out how to make that happen. I’m building a c# web app. I need to upload the files in chunks, and it needs to be transmitted over http (uploading from client to remote cloud server via api). Using a fileupload won’t work, since these files are large (~ 10GB) and the delay between processing the file and hitting the server is way too long. I need to allow the user to type in a file location, hit upload, and stream the file in bytes one chunk at a time.
Any suggestions??

Thank you.
Brian Says:
June 4th, 2011 at 7:54 am
belle and fabiim,
For these situations, you could check out RestSharp: https://github.com/johnsheehan/RestSharp. Its file uploading was based off of this method, but has been extended with asynchronous support (see the ExecuteAsync function), and it might be what you are looking for.
jibin Says:
July 22nd, 2011 at 3:28 am
Hi, i tried the code mentioned here and its not working for me
i got the error or exception
{“The underlying connection was closed: An unexpected error occurred on a receive.”}
{“Unable to read data from the transport connection: An established connection was aborted by the software in your host machine.”}

at the line ” return request.GetResponse() as HttpWebResponse;

I disable antivirus, firewall on my system and still i rxv same error. Any method to solve this error

Thanks
jibin Says:
July 25th, 2011 at 11:09 pm
Hi now i am able to avoid the error
{“The underlying connection was closed: An unexpected error occurred on a receive.”}
{“Unable to read data from the transport connection: An established connection was aborted by the software in your host machine.”}

by properly arrange the header values.(by examine the request via fiddler tool)

But now i am facing this error
The remote server returned an error: (502) Bad Gateway. error , whats the reason?
Dung Says:
July 27th, 2011 at 3:46 am
Excellent and Thank you !
Rheba Charlebois Says:
August 23rd, 2011 at 10:18 pm
Hi there, I found your blog via Google at the same time as looking for a related matter, your site got here up, it looks good. I’ve bookmarked it in my google bookmarks.
Jeff Says:
August 30th, 2011 at 8:26 am
Thanks Brian! After two full days of grinding on this myself I found your post and all was solved. Thanks Again!
Gerben Vos Says:
October 27th, 2011 at 4:39 am
Please use something like
```
String.Format("----------{0:N}", Guid.NewGuid());
```
instead of a hard-coded boundary. For example, it would now be impossible to upload this very page using this code, because the data would contain the boundary. Having a random boundary makes the chance of that happening near-infinitely small.
Melodi Darvin Says:
December 11th, 2011 at 7:11 am
really useful Here’s some mildly amuzing stuff I found: Thought for the day? : I went to the museum where they had all the heads and arms from the statues that are in all the other museums.
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December 30th, 2011 at 2:11 am
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January 3rd, 2012 at 4:23 pm
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Tomas Says:
January 18th, 2012 at 1:45 am
I have searching for solution about three days, the Internet full of codes which show how to post file and additional data using c#. The difference between Your code and others which I found is that Your code actually works. The internet full of junk and it takes a time to find gems :) Thank you Brian!
Eran Says:
February 8th, 2012 at 9:27 am
The code is great, thank you, but I had trouble running it until I applied Pykaso’s fix with the removal of the redundant \r\n before the footer, even though in my form I posted a single file.
The weired thing was that it used to work for me perfectly even without this fix, but then I added some extra parameters to the form and suddenly begun receiving error 500 internal server error.
I found out that the problem was that the last parameter was not a file, and therefore just before the footer this was appended: “–{0}\r\nContent-Disposition: form-data; name=\”{1}\”\r\n\r\n{2}\r\n”
which means an extra \r\n before the footer, and that was my problem! hope it helps…
vittorio Says:
February 9th, 2012 at 11:26 am
GREAT POST!
Pykaso fix for UTF-8 text worked for me.
Tomas Says:
February 10th, 2012 at 1:09 am
Dear developer, are you aware of the Unicode characters bug?

The line

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

is buggy for Unicode characters. Encoding.GetBytes is used to read bytes and postData.Length for length. The length will be different for unicode characters. As Pykaso correctly pointed the Encoding.GetByteCount method should be used for Unicode characters count.
Brian Says:
February 10th, 2012 at 7:04 am
Tomas,
I will update the code now. I was a little worried about making changes that I hadn’t tested but since so many people are mentioning it I changed it. I used .GetByteCount() in 4 different places (“\r\n”, postData, header, and footer). Can you confirm that it is working for you now?

Thanks,
Brian
Brian Says:
February 10th, 2012 at 8:39 am
@Gerben Vos
Good idea with this boundary:
```
String.Format("----------{0:N}", Guid.NewGuid());
```
I have updated the code.
Brian Says:
February 10th, 2012 at 8:43 am
@Pykaso @Eran

I have updated the code to ensure that the extra new line does not get appended at the last field:
```
bool needsCLRF = false;
 
foreach (var param in postParameters)
{
     // Thanks to feedback from commenters, add a CRLF to allow multiple parameters to be added.
     // Skip it on the first parameter, add it to subsequent parameters.
        if (needsCLRF)
            formDataStream.Write(encoding.GetBytes("\r\n"), 0, encoding.GetByteCount("\r\n"));
 
     .....
```
I only add the extra newline at the beginning of each parameter that isn’t the first (I removed the duplicate logic from both file and normal key/value POST. I have tested this locally and confirmed that the extra newline is not sent, but if anyone who was having the issue can confirm that it is fixed for them using the new code, that would be great.
Brian Says:
February 10th, 2012 at 8:44 am
@Stephen
Thanks for the authentication code sample! I have added it in a comment in the code on the post in case anyone else wants to use it.
Dave Says:
April 10th, 2012 at 4:29 pm
Thanks a lot. This helped me a lot.
Derek Says:
April 12th, 2012 at 6:54 am
compiling the code in C# 4 I get an error CS0103: The name ‘Response’ does not exist in the current context.
I’ve made the class a library, and used the library in the main code sections.
It seems this line is the line that is giving a compile time error.

Response.Write(fullResponse);
Brian Says:
May 11th, 2012 at 7:31 pm
Derek,
You probably need to run it like this instead of Response directly.
```
System.Web.HttpContext.Current.Response.Write
```
MMA T-Shirts Says:
May 21st, 2012 at 9:50 pm
Awesome website…

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Jan Says:
August 28th, 2012 at 3:21 am
Excellent post. Good job!
Tomas Says:
October 24th, 2012 at 7:25 am
There is big problem with implementation. The FileParameter object hold Bytes array of file content. If the FormUpload will be used to upload large files and from diferrent thread the chance to get OutOfMemory is very big.
Also Large Object Heap will suffer from such implementation.
Steve Griffiths Says:
November 9th, 2012 at 12:57 pm
Hi Brian,

This looks exactly what I need, but I cannot get it working. I am trying to convert from a cUrl example to c#.

The cUrl example is as follows:
curl -F “file=@myvideo.mp4″ “https://api.mogreet.com/cm/media.upload?client_id=849&token=e15b9a70b18c38fa275496836eab1cb0&type=video&name=VideoPostFromCurl”

(I am sending a jpeg instead)

This is my sample call
private void UploadImage2()
{
OpenFileDialog opendlg = new OpenFileDialog();

if (opendlg.ShowDialog() == DialogResult.OK)
{
// Read file data
FileStream fs = new FileStream(opendlg.FileName, 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”, opendlg.FileName);
postParameters.Add(“fileformat”, “jpg”);
postParameters.Add(“file”, new FormUpload.FileParameter(data, Path.GetFileName(opendlg.FileName), “image/jpeg”));

// Create request and receive response
string strRequest = String.Format(“{0}{1}?client_id={2}&token={3}&type={4}&name={5}”,
_urlPrefix,
“cm/media.upload”,
_clientId,
_token,
“image”,
tbImageName.Text

);

string postURL = strRequest;
string userAgent = “Someone”;
HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);

// Process response
StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
string fullResponse = responseReader.ReadToEnd();
tbResponse.Text += “\r\n\r\n” + fullResponse;
webResponse.Close();

}
}

and this is the response

I am not sure what I am missing – any help appreciated

Thansk

Steve Griffiths

Steve Griffiths Says:
November 9th, 2012 at 1:08 pm

Hi Brian,
My aplogies for formatting the previous comment – this is better

This looks exactly what I need, but I cannot get it working. I am trying to convert from a cUrl example to c#.

The cUrl example is as follows:
curl -F “file=@myvideo.mp4″ “https://api.mogreet.com/cm/media.upload?client_id=849&token=e15b9a70b18c38fa275496836eab1cb0&type=video&name=VideoPostFromCurl”

(I am sending a jpeg instead)

This is my sample call

private void UploadImage2()
        {
            OpenFileDialog opendlg = new OpenFileDialog();
 
            if (opendlg.ShowDialog() == DialogResult.OK)
            {
                // Read file data
                FileStream fs = new FileStream(opendlg.FileName, 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", opendlg.FileName);
                postParameters.Add("fileformat", "jpg");
                postParameters.Add("file", new FormUpload.FileParameter(data, Path.GetFileName(opendlg.FileName), "image/jpeg"));
 
                // Create request and receive response
                string strRequest = String.Format("{0}{1}?client_id={2}&amp;token={3}&amp;type={4}&amp;name={5}",
                                                             _urlPrefix,
                                                             "cm/media.upload",
                                                             _clientId,
                                                             _token,
                                                             "image",
                                                             tbImageName.Text
 
                                                             ); 
 
                string postURL = strRequest;
                string userAgent = "Someone";
                HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);
 
                // Process response
                StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
                string fullResponse = responseReader.ReadToEnd();
                tbResponse.Text += "\r\n\r\n" + fullResponse;
                webResponse.Close();
 
            }
        }
 
        private void UploadImage()
        {
            OpenFileDialog opendlg = new OpenFileDialog();
 
            if (opendlg.ShowDialog() == DialogResult.OK)
            {
 
                FileStream fs = new FileStream(opendlg.FileName, FileMode.Open, FileAccess.Read);
                byte[] data = new byte[fs.Length];
                fs.Read(data, 0, data.Length);
                fs.Close();
 
                string strRequest = String.Format("{0}{1}?client_id={2}&amp;token={3}&amp;name={4}&amp;type={5}&amp;File={6}",
                                                    _urlPrefix,
                                                    "cm/media.upload",
                                                    _clientId,
                                                    _token,
                                                    tbImageName.Text,
                                                    "image",
                                                    data
                                                    );
 
 
                tbResponse.Text = "Request" + "\r\n\r\n" + strRequest + "\r\n\r\n";
                WebRequest request = WebRequest.Create(strRequest);
 
 
                // If required by the server, set the credentials.
                request.Credentials = CredentialCache.DefaultCredentials;
 
                // Get the response.
                HttpWebResponse response = (HttpWebResponse)request.GetResponse();
 
                // Display the status.
                tbResponse.Text += "Response" + "\r\n\r\n" + response.StatusDescription;
 
                // Get the stream containing content returned by the server.
                Stream dataStream = response.GetResponseStream();
 
                // Open the stream using a StreamReader for easy access.
                StreamReader reader = new StreamReader(dataStream);
 
                // Read the content.
                string responseFromServer = reader.ReadToEnd();
 
                // Display the content.
 
                tbResponse.Text += "\r\n\r\n" + responseFromServer;
 
                // Cleanup the streams and the response.
                reader.Close();
                dataStream.Close();
                response.Close();
            }
        }

and this is the response

response status=”error” code=”460″
[CDATA[Asset file name You must include an attachment]]

I am not sure what I am missing – any help appreciated

Thansk

Steve Griffiths

chi Says:
April 4th, 2013 at 12:13 pm
anyone have trouble with uploading a MS word document with images embedded?

Brian Grinstead

Multipart Form Post in C#

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