[移动开发]Android多线程异步通信实例

创建两个线程，实现两个相同的耗时任务

实现2个窗口同时卖火车票；每个窗口卖100张，卖票速度都是1s/张，最后弹出通知票已卖完。

public class MainActivity extends AppCompatActivity {
    private Button button2;
    private Handler handler=new Handler(){
        @Override
        public void handleMessage(@NonNull Message msg) {
            super.handleMessage(msg);
            String s= (String) msg.obj;
            switch (msg.what){
                case 1:
                    Toast.makeText(MainActivity.this,s,Toast.LENGTH_SHORT).show();
                    break;
                case 2:
                    Toast.makeText(MainActivity.this,s,Toast.LENGTH_SHORT).show();
            }
        }
    };
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        button2=findViewById(R.id.bt2);
        button2.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                 int ticket=100;  //一个窗口100张
                new Thread(new Runnable() {
                    @Override
                    public void run() {
                        for (int i=0;i<=ticket;i++){
                            int n=ticket-i-1;
                            Log.i("窗口1","卖掉了1张票，剩余票数："+n);
                            try {
                                Thread.sleep(100);
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        }
                        Message msg=Message.obtain();
                        msg.what=1;
                        msg.obj="窗口一电影票已卖完";
                        handler.sendMessage(msg);
                    }
                }).start();
                new Thread(new Runnable() {
                    @Override
                    public void run() {
                        for (int i=0;i<=ticket;i++){
                            int n=ticket-i-1;
                            Log.i("窗口2","卖掉了1张票，剩余票数："+n);
                            try {
                                Thread.sleep(100);
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        }
                        Message msg=Message.obtain();
                        msg.what=1;
                        msg.obj="窗口二电影票已卖完";
                        handler.sendMessage(msg);
                    }
                }).start();
            }
        });
    }
}