[英]Socket Server controller Android App, Socket error
So basically I want to create some kind of controller for my pc , where on pc, runs java socket server.所以基本上我想为我的电脑创建某种 controller ,在电脑上运行 java 套接字服务器。 The server is working, I have tried it.服务器正常,我试过了。 Also, Port Forwarding in the router is allright.另外,路由器中的端口转发是可以的。 Source code is here :源代码在这里:
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.IOException;
import java.net.ServerSocket;
import java.net.Socket;
import java.util.Scanner;
public class Server {
public static final String ANSI_RESET = "\u001B[0m";
public static final String ANSI_GREEN = "\u001B[32m";
public static final String ANSI_BOLD = "\033[1m";
public static final String ANSI_RED = "\u001B[31m";
public static void main(String[] args) throws IOException {
// Creating server instance
ServerSocket serverSocket = new ServerSocket(2000);
Socket socket = serverSocket.accept();
System.out.println(ANSI_GREEN + "Controller connected" + ANSI_RESET);
// Input Output streams
DataInputStream dataInputStream = new DataInputStream(socket.getInputStream());
DataOutputStream dataOutputStream = new DataOutputStream(socket.getOutputStream());
String command = "";
do {
command = dataInputStream.readUTF();
switch (command.trim()){
case "volume up":
Runtime.getRuntime().exec("cmd /c start src\\sk\\dzurik\\main\\volup.bat");
break;
case "volume down":
Runtime.getRuntime().exec("cmd /c start src\\sk\\dzurik\\main\\voldown.bat");
break;
default:
// Unknown command
break;
}
}while (!command.equals("stop"));
socket.close();
System.out.println(ANSI_RED + "Controller disconnected" + ANSI_RESET);
}
}
And it is supposed to interact with the android app , but I got an error and I can't quite figure out why is that so, here is the source code for ActivityMain :它应该与 android 应用程序交互,但我收到一个错误,我不太明白为什么会这样,这是ActivityMain的源代码:
import androidx.appcompat.app.AppCompatActivity;
import android.content.Context;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.Toast;
import java.io.DataOutputStream;
import java.io.IOException;
import java.net.Socket;
public class MainActivity extends AppCompatActivity {
private boolean connected;
private DataOutputStream dataOutputStream;
private Socket socket;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
// Pripájanie na server
try {
socket = new Socket("172.0.0.1",2000);
dataOutputStream = new DataOutputStream(socket.getOutputStream());
} catch (IOException e) {
Toast.makeText(MainActivity.this, "Not Connected!",
Toast.LENGTH_LONG).show();
}
Button VolumeUp = (Button) findViewById(R.id.VolumeUpButton);
VolumeUp.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
try {
send("volume up");
} catch (IOException e) {
e.printStackTrace();
}
}
});
Button VolumeDown = (Button) findViewById(R.id.VolumeDownButton);
VolumeDown.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
try {
send("volume down");
} catch (IOException e) {
e.printStackTrace();
}
}
});
Button DisconnectButton = (Button) findViewById(R.id.DisconnectButton);
DisconnectButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
try {
send("stop");
} catch (IOException e) {
e.printStackTrace();
}
}
});
}
public void send(String command) throws IOException {
if (socket != null){
dataOutputStream.writeUTF(command);
}
else Toast.makeText(MainActivity.this, "Socket is not defined",
Toast.LENGTH_SHORT).show();
}
public void disconnect() throws IOException {
if (socket != null){
socket.close();
}
else Toast.makeText(MainActivity.this, "Socket is not defined",
Toast.LENGTH_SHORT).show();
}
}
So if you could help me with it, I would be thankful, <3所以,如果你能帮助我,我将不胜感激,<3
Well the reason for that was silly.好吧,这样做的原因很愚蠢。 When I'm loading data from Stream it says nextLine which means the ending point is \n so I could never get that line because I haven't put it there.当我从 Stream 加载数据时,它说 nextLine 这意味着结束点是 \n 所以我永远无法得到那条线,因为我没有把它放在那里。
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