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Transfer a file between Android devices?

I'm making a code and I want to send a mp4 file to another Android device. I have reached to connect both Androids via Wifi and write from one a simple for cycle from 1-20 and the other Android device reads and displays the number that is sent.

Here it is the interesting part of the "sender":

                InetAddress serverAddr = InetAddress.getByName(serverIpAddress);
                Log.d("ClientActivity", "C: Connecting...");
                Socket socket = new Socket(serverAddr, port);
                connected = true;
                while (connected) {
                    try {
                        Log.d("ClientActivity", "C: Sending command.");
                        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(socket
                                    .getOutputStream())), true);


                        for (int i = 1; i < 20; i++) {


                            out.println(i);

                            i=i++;

and the "receiver":

serverSocket = new ServerSocket(SERVERPORT);

                        // listen for incoming clients
                        Socket client = serverSocket.accept();
     BufferedReader in = new BufferedReader(new InputStreamReader(client.getInputStream()),8*1024);

This works great! But I want to send a file from one device to another instead of an int. How can I make this?????

like image 815
Merol Avatar asked Nov 13 '22 16:11

Merol


1 Answers

You need to package the data to the stream through a some kind of data format. One way to do this is to use the common MIME data format which is commonly used to send attachment in email.

I have answered other question related to sending binary via socket using this format in the following SO Question - android add filename to bytestream. You could check the accepted answer for that question.

For your reference, I just copied the code for sending and receiving through socket from that question below.

File  f = new File(path);
BufferedOutputStream out = new BufferedOutputStream( socket.getOutputStream() );
String filename=path.substring(path.lastIndexOf("/")+1);

// create a multipart message
MultipartEntity multipartContent = new MultipartEntity();

// send the file inputstream as data
InputStreamBody isb = new InputStreamBody(new FileInputStream(f), "image/jpeg", filename);

// add key value pair. The key "imageFile" is arbitrary
multipartContent.addPart("imageFile", isb);

multipartContent.writeTo(out);
out.flush();
out.close();

And the code to read it back below using MimeBodyPart that is part of JavaMail.


MimeMultipart multiPartMessage = new MimeMultipart(new DataSource() {
    @Override
    public String getContentType() {
        // this could be anything need be, this is just my test case and illustration
        return "image/jpeg";
    }

    @Override
    public InputStream getInputStream() throws IOException {
        // socket is the socket that you get from Socket.accept()
        BufferedInputStream inputStream = new BufferedInputStream(socket.getInputStream());
        return inputStream;
    }

    @Override
    public String getName() {
        return "socketDataSource";
    }

    @Override
    public OutputStream getOutputStream() throws IOException {
        return socket.getOutputStream();
    }
});

// get the first body of the multipart message
BodyPart bodyPart = multiPartMessage.getBodyPart(0);

// get the filename back from the message
String filename = bodyPart.getFileName();

// get the inputstream back
InputStream bodyInputStream = bodyPart.getInputStream();

// do what you need to do here....
like image 148
momo Avatar answered Dec 24 '22 19:12

momo