Question

我正在编写代码，我想将mp4文件发送到另一台Android设备。我已经通过Wifi连接两个Androids并从一个简单的循环中写入1-20，另一个Android设备读取并显示发送的数字。

这是“发件人”的有趣部分：

                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++;

和“接收者”：

serverSocket = new ServerSocket(SERVERPORT);

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

这很棒！但我想将文件从一个设备发送到另一个设备而不是int。我该如何制作这个?????

Answer 1

您需要通过某种数据格式将数据打包到流中。一种方法是使用常用的MIME数据格式，这种格式通常用于在电子邮件中发送附件。

我已回答了有关在the following SO Question - android add filename to bytestream中使用此格式通过套接字发送二进制文件的其他问题。您可以查看该问题的已接受答案。

供您参考，我刚刚从下面的问题复制了通过套接字发送和接收的代码。

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();

使用作为JavaMail一部分的MimeBodyPart来读取下面的代码。


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....

Answer 2

Google推出了一个开源项目，您可以查看它并大致了解如何在它们之间连接设备和共享文件。

以下是链接：android-fileshare

在Android设备之间传输文件？

2 个答案: