Java 增加游戏难度

我想在我的游戏中添加一个难度，例如，当我单击一个类别（如类别“a”）时，它将显示到下一个布局，用户必须选择一个难度来玩“简单、正常或困难”

对不起，我的句子不好

以下是我的分类：

public class CategoryActivity extends Activity{

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        requestWindowFeature(Window.FEATURE_NO_TITLE);       
        super.onCreate(savedInstanceState);

        setContentView(R.layout.category);

        Button A = (Button) findViewById(R.id.btn_a);
        Button B = (Button) findViewById(R.id.btn_b);
        Button C = (Button) findViewById(R.id.btn_c);
        Button D = (Button) findViewById(R.id.btn_d);
        Button E = (Button) findViewById(R.id.btn_e);
        Button F = (Button) findViewById(R.id.btn_f);
        Button G = (Button) findViewById(R.id.btn_g);
        Button H = (Button) findViewById(R.id.btn_h);
        Button I = (Button) findViewById(R.id.btn_i);
        Button J = (Button) findViewById(R.id.btn_j);

        A.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View arg0) {
                startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

            }
        });

 B.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View arg1) {
                startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));
                }

        });

 C.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg2) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 D.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg3) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 E.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg4) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 F.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg5) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 G.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg6) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 H.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg7) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 I.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg8) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });
 J.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg9) {
            startActivity(new Intent(CategoryActivity.this, DifficultyActivity.class));

        }
    });

    }

---

您应该使用单选按钮，这样用户只能选择一个选项，而不能选择两个或更多

这是一个例子

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附言：你的问题不是很清楚。这使帮助你变得很困难。我希望我的回答可以接受。请参阅手册并执行。

对不起，先生，我想添加代码，但不知道如何添加。。对不起，这不是免费的代码编写服务。您需要做自己的研究，并尝试编写自己的代码@David G.提供了教程的链接。如果你在写你的解决方案时遇到了困难，请随意在上面贴一个新问题。

public void onRadioButtonClicked(View view) {
    // Is the button now checked?
    boolean checked = ((RadioButton) view).isChecked();

    // Check which radio button was clicked
    switch(view.getId()) {
        case R.id.radio_pirates:
            if (checked)
                // Pirates are the best
            break;
        case R.id.radio_ninjas:
            if (checked)
                // Ninjas rule
            break;
    }
}