Question

我有一个countDownTimer，如果用户在第12秒内未点击gameButton我想要调用gameOver方法。问题我或者在countDownTimer为12时立即调用游戏函数，或者计时器只是继续倒计时。所以我试图使用postDelayed（）方法给用户一个完整的秒来点击按钮并让countDownTimer继续，但正如我的代码现在，游戏停在12，无论如何。

import android.app.Activity;
import android.os.CountDownTimer;
import android.os.Handler;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
import android.widget.Toast;



public class GameScreen extends Activity {

    private TextView time;
    private Button start;
    private Button cancel;
    private Button gameButton;
    private CountDownTimer countDownTimer;
    public static int count = 0;
    public static int countFail = 0;

    final Handler handler = new Handler();
    final Runnable r = new Runnable() {
        public void run() {
            handler.postDelayed(this, 1000);
            gameOver();
        }
    };


    private View.OnClickListener btnClickListener = new View.OnClickListener(){

        @Override
        public void onClick(View v) {

            switch(v.getId()){
                case R.id.start_ID :
                    start();
                    break;
                case R.id.cancel :
                    cancel();
                    break;
                case R.id.gameButton_ID :
                    gameButton();
                    break;
            }

        }


    };


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_game_screen);


        start = (Button) findViewById(R.id.start_ID);
        start.setOnClickListener(btnClickListener);
        cancel = (Button) findViewById(R.id.cancel);
        cancel.setOnClickListener(btnClickListener);
        time = (TextView) findViewById(R.id.time);
        gameButton = (Button) findViewById(R.id.gameButton_ID);
        gameButton.setOnClickListener(btnClickListener);


    }

    public void start(){

        time.setText("16");
        //This doesn't work and makes app crash when you hit start button

        countDownTimer = new CountDownTimer(16 * 1000, 1000) {
            @Override
            public void onTick(long millsUntilFinished){
                time.setText("" + millsUntilFinished / 1000);

                //turns textview string to int
                int foo = Integer.parseInt(time.getText().toString());

                if(time.getText().equals("12")){

                    r.run();

                }

            }

            public void onFinish() {
                time.setText("Done !");
            }
        };
        countDownTimer.start();
    }

    private void cancel(){
        if(countDownTimer != null){
            countDownTimer.cancel();
            countDownTimer = null;
        }
    }

    private void gameOver(){
        Toast.makeText(getApplicationContext(), "You scored " + count, Toast.LENGTH_SHORT).show();
        count = 0;
        countFail = 0;
        cancel();
    }

    private void gameButton(){

        int foo = Integer.parseInt(time.getText().toString());

        if(foo  % 2 == 0 ) {
            Toast.makeText(getApplicationContext(), "PASS", Toast.LENGTH_SHORT).show();
            handler.removeCallbacks(r);
            ++count;
        }

        else{
            gameOver();
        }
    }

}

Answer 1

你几乎正在使用postDelayed(Runnable, long)，但不完全正确。我们来看看你的Runnable。

final Runnable r = new Runnable() {
    public void run() {
        handler.postDelayed(this, 1000);
        gameOver();
    }
};

当我们调用r.run();时，它要做的第一件事是告诉你的handler在1000毫秒后运行相同的Runnable，然后调用gameOver()。这实际上会导致你的gameOver()方法被调用两次：一次立即调用，第二次处理程序完成后等待1000毫秒。

相反，您应该将Runnable更改为：

final Runnable r = new Runnable() {
    public void run() {
        gameOver();
    }
};

并称之为：

handler.postDelayed(r, 1000);

希望这有帮助。

Answer 2

下面是我使用的代码，其功能与接受的答案相同，但编写和理解起来非常简单。

final Handler handler = new Handler();
            handler.postDelayed(new Runnable() {
                @Override
                public void run() {
                    //Write whatever to want to do after delay specified (1 sec)
                    Log.d("Handler", "Running Handler");
                }
            }, 1000);

Answer 3

    Thread(Runnable {
        // background work here ... 
        Handler(Looper.getMainLooper()).postDelayed(Runnable {
            // Update UI here
        }, 10000) //it will wait 10 sec before upate ui
    }).start()

Answer 4

不推荐使用无参数的 Handler 构造函数，并且不使用 lambdas 也会使代码看起来很笨拙，所以话虽如此，以下是它看起来更现代的用法：

final Runnable _r_ = new Runnable(){...};

Handler handler = new Handler(Looper.getMainLooper());
handler.postDelayed(() -> _r_.run(), 666);

您也可以更新它以使用方法引用

handler.postDelayed(_r_::run(), 666);

或者更简单的只有一个限定符

handler.postDelayed(_r_, 666);

如何在android studio中正确使用postDelayed（）？

4 个答案: