隐藏随机数在 Java 中确认错误的结果

Question

我对我正在尝试做的事情有一些问题，但最紧迫的是：

我有一个设置隐藏随机数字的按钮，用户必须尝试找出该数字是什么。我当前的代码如下：

设置变量

int max = 9;
int min = 0;
int diff = max-min;
int getNum1;
int a;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_guess);

将 Java 分配给 XML

final TextView urNum1 = findViewById(R.id.txtUrNum1);
final TextView myNum1 = findViewById(R.id.txtMyNum1);
ImageButton up1 = findViewById(R.id.btnUp1);
ImageButton dn1 = findViewById(R.id.btnDn1);
Button chk = findViewById(R.id.btnCheck);
ImageButton start = findViewById(R.id.btnStartGuess);

启动随机发生器

start.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View view) {
        Random rn = new Random();
        urNum1.setText("?");
        a = rn.nextInt(diff+1);
        a += min;
    }
});

检查匹配项

chk.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View view) {
        if (a == 0 && getNum1 == 9) {
            urNum1.setText("0"); }
        else if (a == 1 && getNum1 == 0) {
            urNum1.setText("1"); }
        else if (a == 2 && getNum1 == 1) {
            urNum1.setText("2"); }
        else if (a == 3 && getNum1 == 2) {
            urNum1.setText("3");  }
        else if (a == 4 && getNum1 == 3) {
            urNum1.setText("4"); }
        else if (a == 5 && getNum1 == 4) {
            urNum1.setText("5"); }
        else if (a == 6 && getNum1 == 5) {
            urNum1.setText("6"); }
        else if (a == 7 && getNum1 == 6) {
            urNum1.setText("7"); }
        else if (a == 8 && getNum1 == 7) {
            urNum1.setText("8"); }
        else if (a == 9 && getNum1 == 8) {
            urNum1.setText("9"); }

增加猜测的数字

up1.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                getNum1 = Integer.parseInt(myNum1.getText().toString());
                if (getNum1 == 0) {
                    myNum1.setText("1"); }
                else if (getNum1 == 1 ) {
                    myNum1.setText("2"); }
                else if (getNum1 == 2) {
                    myNum1.setText("3"); }
                else if (getNum1 == 3) {
                    myNum1.setText("4"); }
                else if (getNum1 == 4) {
                    myNum1.setText("5"); }
                else if (getNum1 == 5) {
                    myNum1.setText("6"); }
                else if (getNum1 == 6) {
                    myNum1.setText("7"); }
                else if (getNum1 == 7) {
                    myNum1.setText("8"); }
                else if (getNum1 == 8) {
                    myNum1.setText("9"); }
                else if (getNum1 == 9) {
                    myNum1.setText("0"); }
                });

减少猜测的数字

dn1.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                getNum1 = Integer.parseInt(myNum1.getText().toString());
                if (getNum1 == 0) {
                    myNum1.setText("9"); }
                else if (getNum1 == 9 ) {
                    myNum1.setText("8"); }
                else if (getNum1 == 8) {
                    myNum1.setText("7"); }
                else if (getNum1 == 7) {
                    myNum1.setText("6"); }
                else if (getNum1 == 6) {
                    myNum1.setText("5"); }
                else if (getNum1 == 5) {
                    myNum1.setText("4"); }
                else if (getNum1 == 4) {
                    myNum1.setText("3"); }
                else if (getNum1 == 3) {
                    myNum1.setText("2"); }
                else if (getNum1 == 2) {
                    myNum1.setText("1"); }
                else if (getNum1 == 1) {
                    myNum1.setText("0"); }
            }
        });

我

遇到的问题是，当按下 up1 时，当我检查它们时，数字往往会匹配，但是当我按 dn1 到达匹配的数字时，我选择的数字比匹配的数字低 2。这可能是因为我的编码很复杂，我可能会通过使用 ++ 递增来使其更好，但我似乎也无法使其正常工作。

Answer 1

如果

有人也遇到这个问题，我已经自己找到了解决方案。在我通过递增/递减值来更改 myNum 值的代码中，我只检查"if"语句的每个条件的当前值，而不更改其中的值。所以无论我检查多少次，它总是一次性的。我的修复：

getNum1 = Integer.parseInt(myNum1.getText().toString());
    if (getNum1 == 0) {
        myNum1.setText("9");
        getNum1 = 9; }
    else if (getNum1 == 9 ) {
        myNum1.setText("8");
        getNum1 = 8; }
    else if (getNum1 == 8) {
        myNum1.setText("7");
        getNum1 = 7; }

依此类推。