When you want to post file to Web API / REST client API through c#. You can use below code in C# to post file using Multipart Form Post in C#.
Below is the HTML sample code:
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<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> |
you can post file simply as you are doing in asp.net mvc. After that you can use below C# code:
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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; } } } |
In the above code we have called the function MultipartFormDataPost function with multiple parameters. Below is the implmentation of this function:
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// 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); |
That’s it.
Now you can post file to REST API.
It saved a lot of my time. Awesome article. Can you please tell me how to pass the array in the REST API as a parameter for POST function. Thank you
Amazing code! Thanks!