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我想在 2 个 android 设备之间建立通信。使用服务器到客户端（通过服务器）方法。所以通常我在考虑将文件发送到服务器（pc）然后在服务器上获取文件并发送到另一个设备。所以我正在工作第二部分，试图将文件从 pc 发送到 android。但由于某种原因，客户端无法连接到服务器。这是我的代码；

public class TCPServer extends Thread {

public static final int SERVERPORT = 8901;

public static void main() {

    try {
        System.out.println("S: Connecting...");

        ServerSocket serverSocket = new ServerSocket(SERVERPORT);
        System.out.println("S: Socket Established...");

        Socket client = serverSocket.accept();
        System.out.println("S: Receiving...");

        ObjectOutputStream put = new ObjectOutputStream(
                client.getOutputStream());

        String s = "adios.wav";
        String str = "C:/";
        String path = str + s;
        System.out.println("The requested file is path: " + path);
        System.out.println("The requested file is : " + s);
        File f = new File(path);
        if (f.isFile()) {
            FileInputStream fis = new FileInputStream(f);

            byte[] buf = new byte[1024];
            int read;
            while ((read = fis.read(buf, 0, 1024)) != -1) {
                put.write(buf, 0, read);
                put.flush();
            }

            System.out.println("File transfered");
            client.close();
            serverSocket.close();
            fis.close();
        }

    } catch (Exception e) {
        System.out.println("S: Error");
        e.printStackTrace();
    } finally {

    }
}
}

和客户；

public class MainActivity extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    Socket s = null;
    BufferedInputStream get = null;

    try {
        s = new Socket("192.168.198.1", 8901);
        get = new BufferedInputStream(s.getInputStream());

        int u;

        String str = "/mnt/sdcard/ad.wav";
        FileOutputStream fs = new FileOutputStream(new File(str));
        byte jj[] = new byte[1024];
        while ((u = get.read(jj, 0, 1024)) != -1) {
            fs.write(jj, 0, u);
        }
        fs.close();
        System.out.println("File received");
        s.close();
    } catch (Exception e) {
        e.printStackTrace();
        System.exit(0);
    }
}
}
4

1 回答 1

2

我认为客户端部分不会起作用。

您应该在另一个线程或 AsyncTask 中执行套接字的工作。

例如：

public class MainActivity extends Activity {

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        ...
        new doitAsync().execute();
    }

   class doitAsync extends AsyncTask<Void, Integer, Integer>
   {

       @Override
       protected void onPostExecute(Integer result)
       {
           if (result == -1)
           {
               System.exit(0);
           }
       }

       @Override
       protected Integer doInBackground(Void... params)
       {
           Socket s = null;
           BufferedInputStream get = null;

           try {
               s = new Socket("192.168.198.1", 8901);
               get = new BufferedInputStream(s.getInputStream());

               int u;

               String str = "/mnt/sdcard/ad.wav";
               FileOutputStream fs = new FileOutputStream(new File(str));
               byte jj[] = new byte[1024];
               while ((u = get.read(jj, 0, 1024)) != -1) {
                   fs.write(jj, 0, u);
               }
               fs.close();
               System.out.println("File received");
               s.close();
           } catch (Exception e) {
               e.printStackTrace();
               return -1; // failed
           }
           return 0;
       }

此外，您可以考虑

onProgressUpdate()

如果你想更新一个进度条，那么应用程序就不会被冻结。

于 2013-05-12T17:42:40.580 回答