c# - Post a file using HTTPWebrequest (multipart/form-data) gives an error (A potentially dangerous Request.Form value was detected from the client) -

i want send file using http post mvc web application. file contains html tags. below code have tried.

public void postmultiplefiles(string url, string[] files) {     string boundary = "----------------------------" + datetime.now.ticks.tostring("x");     httpwebrequest httpwebrequest = (httpwebrequest)webrequest.create(url);     httpwebrequest.contenttype = "multipart/form-data; boundary=" + boundary;     httpwebrequest.method = "post";     httpwebrequest.keepalive = true;     httpwebrequest.credentials = system.net.credentialcache.defaultcredentials;     stream memstream = new system.io.memorystream();     byte[] boundarybytes =system.text.encoding.ascii.getbytes("\r\n--" + boundary     +"\r\n");     string formdatatemplate = "\r\n--" + boundary + "\r\ncontent-disposition:  form-data; name=\"{0}\";\r\n\r\n{1}";     string headertemplate = "content-disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n content-type: application/octet-stream\r\n\r\n";     memstream.write(boundarybytes, 0, boundarybytes.length);     (int = 0; < files.length; i++)     {         string header = string.format(headertemplate, "file" + i, files[i]);         //string header = string.format(headertemplate, "uplthefile", files[i]);         byte[] headerbytes = system.text.encoding.utf8.getbytes(header);         memstream.write(headerbytes, 0, headerbytes.length);         filestream filestream = new filestream(files[i], filemode.open,         fileaccess.read);         byte[] buffer = new byte[1024];         int bytesread = 0;         while ((bytesread = filestream.read(buffer, 0, buffer.length)) != 0)         {             memstream.write(buffer, 0, bytesread);         }         memstream.write(boundarybytes, 0, boundarybytes.length);         filestream.close();     }     httpwebrequest.contentlength = memstream.length;     stream requeststream = httpwebrequest.getrequeststream();     memstream.position = 0;     byte[] tempbuffer = new byte[memstream.length];     memstream.read(tempbuffer, 0, tempbuffer.length);     memstream.close();     requeststream.write(tempbuffer, 0, tempbuffer.length);     requeststream.close();     try     {         webresponse webresponse = httpwebrequest.getresponse();         stream stream = webresponse.getresponsestream();         streamreader reader = new streamreader(stream);         string var = reader.readtoend();      }     catch (exception ex)     {         response.innerhtml = ex.message;     }     httpwebrequest = null; } 

but when request received mvc application, exception occured. exception - "a potentially dangerous request.form value detected client". reason exception , how can post file?

you can decorate action allowhtml attribute:

[allowhtml] public void postmultiplefiles(string url, string[] files) 

note potential security risk! see this msdn article more info.