I am very new to java and servlet programming. I am not sure whether it is possible to write a servlet which when passed a URL from the local client machine, uploads the file to the server.
basically on the client machine we have a C# program and on the server side we have Apache-tomcat installed. I need to upload file(s) to the server using C# program on client machine.
Should I provide any more information (?)
Thanks in Advance
Note this code illustrates the general idea and not guaranteed to work without modification. The C# file upload part
// this code shows you how the browsers wrap the file upload request, you still can fine a way simpler code to do the same thing.
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);
for (int i = 0; i < 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;
}
and to understand how the above code was written you might wanna take a look at How does HTTP file upload work?
POST /upload?upload_progress_id=12344 HTTP/1.1
Host: localhost:3000
Content-Length: 1325
Origin: http://localhost:3000
... other headers ...
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryePkpFF7tjBAqx29L
------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="MAX_FILE_SIZE"
100000
------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="uploadedfile"; filename="hello.o"
Content-Type: application/x-object
... contents of file goes here ...
------WebKitFormBoundaryePkpFF7tjBAqx29L--
and finally all you have to do is to implement a servlet that can handle the file upload request, then you do whatever that you want to do with the file, take a look at this file upload tutorial
protected void processRequest(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
// Create path components to save the file
final String path = request.getParameter("destination");
final Part filePart = request.getPart("file");
final String fileName = getFileName(filePart);
OutputStream out = null;
InputStream filecontent = null;
final PrintWriter writer = response.getWriter();
try {
out = new FileOutputStream(new File(path + File.separator
+ fileName));
filecontent = filePart.getInputStream();
int read = 0;
final byte[] bytes = new byte[1024];
while ((read = filecontent.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
writer.println("New file " + fileName + " created at " + path);
LOGGER.log(Level.INFO, "File{0}being uploaded to {1}",
new Object[]{fileName, path});
} catch (FileNotFoundException fne) {
writer.println("You either did not specify a file to upload or are "
+ "trying to upload a file to a protected or nonexistent "
+ "location.");
writer.println("<br/> ERROR: " + fne.getMessage());
LOGGER.log(Level.SEVERE, "Problems during file upload. Error: {0}",
new Object[]{fne.getMessage()});
} finally {
if (out != null) {
out.close();
}
if (filecontent != null) {
filecontent.close();
}
if (writer != null) {
writer.close();
}
}
}
private String getFileName(final Part part) {
final String partHeader = part.getHeader("content-disposition");
LOGGER.log(Level.INFO, "Part Header = {0}", partHeader);
for (String content : part.getHeader("content-disposition").split(";")) {
if (content.trim().startsWith("filename")) {
return content.substring(
content.indexOf('=') + 1).trim().replace("\"", "");
}
}
return null;
}
