Question

我尝试使用JScript在Android Studio上创建随机数生成器。该应用程序可以正常工作，但是当用户输入任何内容时它会崩溃，并且在用户重新启动应用程序之前它也不会更改随机数。我尝试过以下代码：

    public class MainActivity extends AppCompatActivity {
        int randomNumber;
        public void checkGuess(View view){
            EditText guessedNumber = (EditText) findViewById(R.id.guessedNumber);
            String guessedNumberString = guessedNumber.getText().toString();
            int guessedNumberInt = Integer.parseInt(guessedNumberString);
            String message = "";

            if (guessedNumberInt>randomNumber){
                message = "Too large!";
            }else if(guessedNumberInt<randomNumber){
                message = "Too small!";
            }else {
                message = "correct!";
            }

            Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
}


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

        Random randomGenerator = new Random();
        int randomNumber = randomGenerator.nextInt(21);
    }
}

当用户猜对时，我应该怎么做才能更改随机数，以及当用户什么都不输入时如何修复应用程序崩溃？

Answer 1

使随机randomGenerator全局化。问题是你只生成一次随机数.onCreate只会被调用。你需要在每次调用时创建一个随机数。

像这样更改onCreate

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

        randomGenerator = new Random();

    }
}

然后像这样改变你的checkGuess方法

        public void checkGuess(View view){
            int randomNumber = randomGenerator.nextInt(21);
            EditText guessedNumber = (EditText) findViewById(R.id.guessedNumber);
            String guessedNumberString = guessedNumber.getText().toString();
           if(TextUtils.isEmpty(guessedNumberString)){
            Toast.makeText(getApplicationContext(), "Please enter a number", Toast.LENGTH_LONG).show();
            return;
           }
            int guessedNumberInt = Integer.parseInt(guessedNumberString);
            String message = "";



            if (guessedNumberInt>randomNumber){
                message = "Too large!";
            }else if(guessedNumberInt<randomNumber){
                message = "Too small!";
            }else {
                message = "correct!";
            }

            Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
}

Answer 2

这里用它来解决你的问题：

public void checkGuess(View view){
            EditText guessedNumber = (EditText) findViewById(R.id.guessedNumber);

            String guessedNumberString = guessedNumber.getText().toString();
        String message = "";
            int guessedNumberInt =0;
            if (guessedNumberString.length() > 0){
                Random randomGenerator = new Random();
                int randomNumber = randomGenerator.nextInt(21);
                guessedNumberInt = Integer.parseInt(guessedNumberString);
                if (guessedNumberInt>randomNumber){
                    message = "Too large!";
                }else if(guessedNumberInt<randomNumber){
                    message = "Too small!";
                }else {
                    message = "correct!";
                }

                Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
            }
            }else{
                //show message to enter a value
            }
    }

Android猜谜游戏？如何在用户正确使用时更改随机数并在用户输入任何内容时解决崩溃问题？

2 个答案: