Multipart forms from C# client
Thanks for the answers, everybody! I recently had to get this to work, and used your suggestions heavily. However, there were a couple of tricky parts that did not work as expected, mostly having to do with actually including the file (which was an important part of the question). There are a lot of answers here already, but I think this may be useful to someone in the future (I could not find many clear examples of this online). I wrote a blog post that explains it a little more.
Basically, I first tried to pass in the file data as a UTF8 encoded string, but I was having problems with encoding files (it worked fine for a plain text file, but when uploading a Word Document, for example, if I tried to save the file that was passed through to the posted form using Request.Files[0].SaveAs(), opening the file in Word did not work properly. I found that if you write the file data directly using a Stream (rather than a StringBuilder), it worked as expected. Also, I made a couple of modifications that made it easier for me to understand.
By the way, the Multipart Forms Request for Comments and the W3C Recommendation for mulitpart/form-data are a couple of useful resources in case anyone needs a reference for the specification.
I changed the WebHelpers class to be a bit smaller and have simpler interfaces, it is now called FormUpload. If you pass a FormUpload.FileParameter you can pass the byte[] contents along with a file name and content type, and if you pass a string, it will treat it as a standard name/value combination.
Here is the FormUpload class:
// 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 calling code, which uploads a file and a few normal post parameters:
// 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);
This is cut and pasted from some sample code I wrote, hopefully it should give the basics. It only supports File data and form-data at the moment.
public class PostData
{
private List<PostDataParam> m_Params;
public List<PostDataParam> Params
{
get { return m_Params; }
set { m_Params = value; }
}
public PostData()
{
m_Params = new List<PostDataParam>();
// Add sample param
m_Params.Add(new PostDataParam("email", "MyEmail", PostDataParamType.Field));
}
/// <summary>
/// Returns the parameters array formatted for multi-part/form data
/// </summary>
/// <returns></returns>
public string GetPostData()
{
// Get boundary, default is --AaB03x
string boundary = ConfigurationManager.AppSettings["ContentBoundary"].ToString();
StringBuilder sb = new StringBuilder();
foreach (PostDataParam p in m_Params)
{
sb.AppendLine(boundary);
if (p.Type == PostDataParamType.File)
{
sb.AppendLine(string.Format("Content-Disposition: file; name=\"{0}\"; filename=\"{1}\"", p.Name, p.FileName));
sb.AppendLine("Content-Type: text/plain");
sb.AppendLine();
sb.AppendLine(p.Value);
}
else
{
sb.AppendLine(string.Format("Content-Disposition: form-data; name=\"{0}\"", p.Name));
sb.AppendLine();
sb.AppendLine(p.Value);
}
}
sb.AppendLine(boundary);
return sb.ToString();
}
}
public enum PostDataParamType
{
Field,
File
}
public class PostDataParam
{
public PostDataParam(string name, string value, PostDataParamType type)
{
Name = name;
Value = value;
Type = type;
}
public string Name;
public string FileName;
public string Value;
public PostDataParamType Type;
}
To send the data you then need to:
HttpWebRequest oRequest = null;
oRequest = (HttpWebRequest)HttpWebRequest.Create(oURL.URL);
oRequest.ContentType = "multipart/form-data";
oRequest.Method = "POST";
PostData pData = new PostData();
byte[] buffer = encoding.GetBytes(pData.GetPostData());
// Set content length of our data
oRequest.ContentLength = buffer.Length;
// Dump our buffered postdata to the stream, booyah
oStream = oRequest.GetRequestStream();
oStream.Write(buffer, 0, buffer.Length);
oStream.Close();
// get the response
oResponse = (HttpWebResponse)oRequest.GetResponse();
Hope thats clear, i've cut and pasted from a few sources to get that tidier.
With .NET 4.5 you currently could use System.Net.Http namespace. Below the example for uploading single file using multipart form data.
using System;
using System.IO;
using System.Net.Http;
namespace HttpClientTest
{
class Program
{
static void Main(string[] args)
{
var client = new HttpClient();
var content = new MultipartFormDataContent();
content.Add(new StreamContent(File.Open("../../Image1.png", FileMode.Open)), "Image", "Image.png");
content.Add(new StringContent("Place string content here"), "Content-Id in the HTTP");
var result = client.PostAsync("https://hostname/api/Account/UploadAvatar", content);
Console.WriteLine(result.Result.ToString());
}
}
}