Question

我正在尝试创建一个充当Web Socket服务器的Android应用程序。这是我的MainActivity.java。

package com.example.websocket;

import java.net.InetSocketAddress;
import java.net.UnknownHostException;

import org.apache.http.conn.util.InetAddressUtils;
import org.java_websocket.WebSocket;
import org.java_websocket.handshake.ClientHandshake;
import org.java_websocket.server.WebSocketServer;

import android.os.Bundle;
import android.app.Activity;
import android.view.Menu;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;

public class MainActivity extends Activity {

    EditText port, msg;
    Button listen, send;
    TextView status;
    int p;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        port = (EditText)findViewById(R.id.editText1);
        msg = (EditText)findViewById(R.id.editText2);
        listen = (Button)findViewById(R.id.button1);
        send = (Button)findViewById(R.id.button2);
        status = (TextView)findViewById(R.id.textView1);

        listen.setOnClickListener(new View.OnClickListener() 
        {   
            @Override
            public void onClick(View arg0) {
                String temp = port.getText().toString();
                p = Integer.parseInt(temp);
                try 
                {
                    custom_web_socket wsocket = new custom_web_socket(p);
                    wsocket.start();
                } 
                catch (UnknownHostException e) {
                    e.printStackTrace();
                }
            }
        });
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }

    public class custom_web_socket extends WebSocketServer
    {

        public custom_web_socket(int port_add) throws UnknownHostException {
            super(new InetSocketAddress(port_add));
            // pass
        }

        @Override
        public void onClose(WebSocket arg0, int arg1, String arg2, boolean arg3) {
            status.setText("Connection closed.");

        }

        @Override
        public void onError(WebSocket arg0, Exception arg1) {
            status.setText((CharSequence) arg1);

        }

        @Override
        public void onMessage(WebSocket arg0, String arg1) {
            // TODO Auto-generated method stub

        }

        @Override
        public void onOpen(WebSocket arg0, ClientHandshake arg1) {
            status.setText("Connected");
        }

    }

}

运行程序时，应用程序启动。但是，当我尝试在EditText字段中输入值时，应用程序会挂起并结束。我将此示例称为Web套接字 - https://github.com/TooTallNate/Java-WebSocket/blob/master/src/main/example/ChatServer.java

我已经导入了必要的jar文件。这是我第一次尝试android编程，我不确定问题是什么。

Answer 1

您正在使用onClick事件。因此，无论何时尝试输入内容，都会调用您的方法。并挂断电话。

当用户输入完代码后，您的操作就会发生。您可以使用以下代码

port.setOnEditorActionListener(
new EditText.OnEditorActionListener() {
 @Override 
  public boolean onEditorAction(TextView v, int actionId, KeyEvent   
       event)      {
      if (actionId == EditorInfo.IME_ACTION_SEARCH ||
                    actionId == EditorInfo.IME_ACTION_DONE ||
                         event.getAction() == KeyEvent.ACTION_DOWN &&

                     event.getKeyCode()==KeyEvent.KEYCODE_ENTER)        
                {
                     if (!event.isShiftPressed()) {
                                // the user is done typing. 

                               return true; // consume.
                         }                
                  }
               return false; // pass on to other listeners. 
            }
         });

另请参阅http://developer.android.com/reference/android/widget/EditText.html 你会得到清晰的想法。

Android程序突然结束

1 个答案: