Introduction
Okay, so you like the idea behind WebClient
's UploadFile
function, but you can't get it to do all the things you want? That's were I was at a week ago. This is what I learned in the past week. UploadFile
seems good enough on the surface but you can run into some tricky things fast, like if you have some cookies to manage or you want to specify the content type, or change the name of the file input in the form? Well, this function will take care of those problems and should make simulating a web form with input type=file
as easy as pi, er.. pie. Under the hood is a HttpWebRequest
version of UploadFile
that you could use to add additional functionality.
For example...
You would like to simulate this web form from your C# application:
<form action ="http://localhost/test.php" method = POST>
<input type = text name = uname>
<input type = password name =passwd>
<input type = FILE name = uploadfile>
<input type=submit>
</form>
You could run UploadFileEx
like this:
CookieContainer cookies = new CookieContainer();
NameValueCollection querystring = new NameValueCollection();
querystring["uname"]="uname";
querystring["passwd"]="snake3";
string uploadfile;
uploadfile = "c:\\test.jpg";
string outdata = UploadFileEx(uploadfile,
"http://localhost/test.php","uploadfile", "image/pjpeg",
querystring,cookies);
So on to the code...
So, this is UploadFileEx
:
public static string UploadFileEx( string uploadfile, string url,
string fileFormName, string contenttype,NameValueCollection querystring,
CookieContainer cookies)
{
if( (fileFormName== null) ||
(fileFormName.Length ==0))
{
fileFormName = "file";
}
if( (contenttype== null) ||
(contenttype.Length ==0))
{
contenttype = "application/octet-stream";
}
string postdata;
postdata = "?";
if (querystring!=null)
{
foreach(string key in querystring.Keys)
{
postdata+= key +"=" + querystring.Get(key)+"&";
}
}
Uri uri = new Uri(url+postdata);
string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest webrequest = (HttpWebRequest)WebRequest.Create(uri);
webrequest.CookieContainer = cookies;
webrequest.ContentType = "multipart/form-data; boundary=" + boundary;
webrequest.Method = "POST";
StringBuilder sb = new StringBuilder();
sb.Append("--");
sb.Append(boundary);
sb.Append("\r\n");
sb.Append("Content-Disposition: form-data; name=\"");
sb.Append(fileFormName);
sb.Append("\"; filename=\"");
sb.Append(Path.GetFileName(uploadfile));
sb.Append("\"");
sb.Append("\r\n");
sb.Append("Content-Type: ");
sb.Append(contenttype);
sb.Append("\r\n");
sb.Append("\r\n");
string postHeader = sb.ToString();
byte[] postHeaderBytes = Encoding.UTF8.GetBytes(postHeader);
byte[] boundaryBytes =
Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
FileStream fileStream = new FileStream(uploadfile,
FileMode.Open, FileAccess.Read);
long length = postHeaderBytes.Length + fileStream.Length +
boundaryBytes.Length;
webrequest.ContentLength = length;
Stream requestStream = webrequest.GetRequestStream();
requestStream.Write(postHeaderBytes, 0, postHeaderBytes.Length);
byte[] buffer = new Byte[checked((uint)Math.Min(4096,
(int)fileStream.Length))];
int bytesRead = 0;
while ( (bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0 )
requestStream.Write(buffer, 0, bytesRead);
requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
WebResponse responce = webrequest.GetResponse();
Stream s = responce.GetResponseStream();
StreamReader sr = new StreamReader(s);
return sr.ReadToEnd();
}
This might help too..
You can handle the post in various ways but this is the PHP file I wrote to test:
<?php
print_r($_REQUEST);
$uploadDir = '%SOMEPATH';
$uploadFile = $uploadDir . $_FILES['userfile']['name'];
print "<PRE>";
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadFile))
{
print "File is valid, and was successfully uploaded. ";
}
else
{
print "Possible file upload attack! Here's some debugging info:\n";
print_r($_FILES);
}
print "</PRE>";
?>
Credits...
I found most of this out by trial and error searching the web. One major source of this information was this article.