Question

我有简单的Java服务器：

public class Main {

    public static void main(String[] args) throws IOException {
        System.out.println("Welcome to Server side");
        BufferedReader in;
        PrintWriter out;

        ServerSocket servers = null;
        Socket fromClient = null;

        // create server socket
        try {
            servers = new ServerSocket(4444);
        } catch (IOException e) {
            System.out.println("Couldn't listen to port 4444");
            System.exit(-1);
        }

        try {
            System.out.print("Waiting for a client...");
            fromClient = servers.accept();
            System.out.println("Client connected");
        } catch (IOException e) {
            System.out.println("Can't accept");
            System.exit(-1);
        }

        in = new BufferedReader(new InputStreamReader(fromClient.getInputStream()));
        out = new PrintWriter(fromClient.getOutputStream(), true);
        String input;

        System.out.println("Wait for messages");
        while ((input = in.readLine()) != null) {
            if (input.equalsIgnoreCase("exit")) break;
            out.println("S ::: " + input);
            System.out.println(input);
        }
        out.close();
        in.close();
        fromClient.close();
        servers.close();
    }
}

和简单的Android客户端：

@Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        editText = (EditText) findViewById(R.id.editText);
        button = (Button) findViewById(R.id.button);
        button.setOnClickListener(this);
        new MyAsync().execute();
    }
---------------------
public class MyAsync extends AsyncTask<Void, Void, Void>{

        @Override
        protected Void doInBackground(Void... params) {
            try {
                fromServer = new Socket("192.168.0.103",4444);
                in = new BufferedReader(new InputStreamReader(fromServer.getInputStream()));
                out = new PrintWriter(fromServer.getOutputStream(), true);
            } catch (IOException e) {
                e.printStackTrace();
            }
            return null;
        }
    }

-------------------

 @Override
    public void onClick(View v) {
        if (v.getId() == R.id.button){
            out.println(editText.getText().toString());
        }
    }

一切顺利。我从Android发送消息到服务器并在服务器中打印此消息。但我想发送Object，例如User：

public class User {
    private int age;
    private String fio;

    public User() {
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getFio() {
        return fio;
    }

    public void setFio(String fio) {
        this.fio = fio;
    }

在Android中我可以写：

User user = new User();
out.print(user);

但我不明白我怎么能在服务器端读到这个？

Answer 1

您无法使用print()执行此操作。在发件人处使用ObjectOutputStream.writeObject()，在接收者处使用ObjectInputStream.readObject()。您需要将User课程调整为implement Serializable并提供private static long serialVersionUID值。

注意不要写这样的代码。取决于try块中代码成功的代码应位于同一try块内。

如何通过套接字从android传递和接收对象到Java服务器？

1 个答案: