Question

在游戏中输出随机数（loadG1）并显示4秒。一旦四秒钟结束，它就会消失，用户必须输入其值才能获得一个点。一旦用户按下回车键，他们的输入框就会消失，程序必须等到更长的CountDownTimer（当前为18秒）结束才能看到他们的分数。到目前为止，他们只能获得一分。我想要发生的是，一旦他们输入答案（无论是否正确），重复4秒倒数计时器的内容。虽然，我想在倒数计时器再次启动时也输出loadG1。所以他们输入一个答案，一个新的随机loadG1再显示4秒然后消失，他们输入另一个答案等等......直到更长的计时器结束。我该怎么做呢？我知道循环不起作用，因为我尝试了它只是崩溃了我的应用程序。如果有人可以帮我这样做，我会非常感激，并告诉我如何实现代码。注意：我是学习Java和Android开发人员的新手。非常感谢提前。

以下是代码：

public class game1 extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_game1);


final Button loseStarter1;

    loseStarter1 = (Button) findViewById(R.id.Starter1);
    loseStarter1.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            loseStarter1.setVisibility(View.GONE);

            final int[] score = {0};
            Random generateG1 = new Random();
            final int loadG1 = generateG1.nextInt(1000000)+10000;
            final TextView number = (TextView) findViewById(R.id.number);
            number.setText(" "+loadG1);

            new CountDownTimer(18000, 1000) {
                @Override
                public void onTick (long millisUntilFinished) {
                }
                public void onFinish() {
                    TextView result = (TextView) findViewById(R.id.outcome);
                    result.setText("Score: "+ score[0]);
                    TextView prompt = (TextView) findViewById(R.id.prompt);
                    prompt.setVisibility(View.GONE);
                }
            }.start();


                    new CountDownTimer(4000, 1000) {

                        @Override
                        public void onTick(long millisUntilFinished) {

                        }

                        @Override
                        public void onFinish() {
                            number.setVisibility(View.GONE);
                            final TextView prompt = (TextView) findViewById(R.id.prompt);
                            prompt.setText(" Enter the number");
                            final EditText input = (EditText) findViewById(R.id.enterAnswer);
                            input.setVisibility(View.VISIBLE);
                            input.setOnKeyListener(new View.OnKeyListener() {
                                @Override
                                public boolean onKey(View v, int keyCode, KeyEvent event) {
                                    if (event.getAction() == KeyEvent.ACTION_DOWN) {
                                        switch (keyCode) {
                                            case KeyEvent.KEYCODE_ENTER:
                                                Editable answer = input.getText();
                                                int finalAnswer = Integer.parseInt(String.valueOf(answer));
                                                int finalLoadG1 = Integer.parseInt(String.valueOf(loadG1));
                                                input.setVisibility(View.GONE);
                                                prompt.setVisibility(View.GONE);
                                                if (finalAnswer == finalLoadG1) {
                                                    score[0]++;
                                                }

                                                return true;
                                            default:
                                        }
                                    }
                                    return false;
                                }
                            });
                        }
                    }.start();

            };
        });
    }

}

Answer 1

RESOLVED

I put the following after the score[0]++:

number.setVisibility(View.VISIBLE);
final int loadG1 = generateG1.nextInt(1000000) + 10000;
number.setText(" " + loadG1);
input.getText().clear();

start();

This way, all necessary changes are made and the timer begins anew. I also modified the visibility of the prompt TextView to ensure it only appear when the user is able to input an answer. Took me a while but this has been resolved.

如何以特定方式重复此倒数计时器......？

1 个答案: