java - 分数变量不会更新，并且必须是最终的？

Question

我遇到的问题是我的程序中的分数计数器变量“score”不能高于 1。如果用户输入的值等于显示的随机数 (LoadG1)，他们将获得一分。然后在最后输出，如较长的计时器所示。在较短的计时器中，分数被添加。 int 变量在开头声明。全部如下所示。现在我知道它是最终的，这很可能是导致我的问题的原因，但 IDE 要求它是最终的，否则我只能在一种方法中调用变量“分数”(在我的例子中是倒计时器)。这给我带来了问题。如果用户每次输入正确，我打算在每次 4 秒计时器重复时添加一个点，尽管目前它不能超过 1。我希望最后输出最终分数，如下所示。

代码:

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

final CountDownTimer loop = 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.setVisibility(View.VISIBLE);
    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]++;
                        }

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

                        start();

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

new CountDownTimer(24000, 1000) {
@Override
public void onTick (long millisUntilFinished) {
}
@Override
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);
    final EditText input = (EditText) findViewById(R.id.enterAnswer);
    input.setVisibility(View.GONE);
    loop.cancel();
}
}.start();

如果有人能够解决我的问题，我将不胜感激，提前致谢。

Answer 1

因为您没有读取生成的随机值，而只是读取第一个值。所以第一次答案是正确的，但下次你的答案不同，但 if 与第一次生成的随机数进行比较，所以它不计为有效答案(如果你输入相同的第一个数字所有次数就会计数)。因为有当前号码，所以每次都需要从TextEdit号码中读取号码。

所以可能是:

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

final CountDownTimer loop = 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.setVisibility(View.VISIBLE);
    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));
                        // here we get from text field the current correct value
                        int finalLoadG1 = Integer.parseInt(String.valueOf(number.getText()));
                        input.setVisibility(View.GONE);
                        prompt.setVisibility(View.GONE);
                        if (finalAnswer == finalLoadG1) {
                            score[0]++;
                        }

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

                        start();

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

new CountDownTimer(24000, 1000) {
@Override
public void onTick (long millisUntilFinished) {
}
@Override
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);
    final EditText input = (EditText) findViewById(R.id.enterAnswer);
    input.setVisibility(View.GONE);
    loop.cancel();
}
}.start();

此外，您可以(并且应该)使用 AtomicInteger 而不是 int[]，您感兴趣的方法是:

int AtomicInteger#get();
int AtomicInteger#incrementAndGet()

因此将分数声明为 AtomicInteger，如下所示:

final AtomicInteger score = new AtomicInteger();

然后代替 score[0]++; 执行 score.incrementAndGet();

然后当你读取结果时:score.get();