Reputation: 3911
hey guys, m using an api of "Bits on the Run" following is the code of upload API
public string Upload(string uploadUrl, NameValueCollection args, string filePath)
{
_queryString = args; //no required args
WebClient client = createWebClient();
_queryString["api_format"] = APIFormat ?? "xml"; //xml if not specified - normally set in required args routine
queryStringToArgs();
string callUrl = _apiURL + uploadUrl + "?" + _args;
callUrl = uploadUrl + "?" + _args;
try {
byte[] response = client.UploadFile(callUrl, filePath);
return Encoding.UTF8.GetString(response);
} catch {
return "";
}
}
and below is my code to upload a file, m using FileUpload control to get the full path of a file(but m not succeeded in that)...
botr = new BotR.API.BotRAPI("key", "secret_code");
var response = doc.Descendants("link").FirstOrDefault();
string url = string.Format("{0}://{1}{2}", response.Element("protocol").Value, response.Element("address").Value, response.Element("path").Value);
//here i want fullpath of the file, how can i achieve that here
string filePath = fileUpload.PostedFile.FileName;//"C://Documents and Settings//rkrishna//My Documents//Visual Studio 2008//Projects//BitsOnTheRun//BitsOnTheRun//rough_test.mp4";
col = new NameValueCollection();
FileStream fs = new FileStream(filePath, FileMode.Open);
col["file_size"] = fs.Length.ToString();
col["file_md5"] = BitConverter.ToString(HashAlgorithm.Create("MD5").ComputeHash(fs)).Replace("-", "").ToLower();
col["key"] = response.Element("query").Element("key").Value;
col["token"] = response.Element("query").Element("token").Value;
fs.Dispose();
string uploadResponse = botr.Upload(url, col, filePath);
i read in some forums saying that for some security purpose you can't get fullpath of a file from client side. If it is true then how can i achieve file upload in my scenario ?
Upvotes: 0
Views: 3952
Reputation: 4400
Yes, this is true, for security reason you cannot get the fullpath of the client machine, what you can do is, try the following,
Stream stream = fileUpload.PostedFile.InputStream;
stream.Read(bytes, 0, fileUpload.PostedFile.ContentLength);
instead of creating your own FileStream use the stream provided by the FileUploadControl. Hoep it shall help.
Upvotes: 1