Question

我想要做的就是这样做，以便当通过一些猜测时，它会使它显示相同的消息，直到它们关闭并重新进入应用程序但我无法弄清楚如何这样做。

我目前的游戏代码是：

    String randtext ="";
      Random rand = new Random();
      int noofguess = 0;
        int n = rand.nextInt(20)+1; //generate random number
        int userguess = Integer.parseInt(etCommentInput.getText().toString());
        if (userguess < 1 || userguess > 20){
            tvCommentOuput.setText("Please enter a number between 1-20!");
        }
        else if (userguess == n){
            tvCommentOuput.setText("You Got It Right!");
        } else if (userguess > n) {
            tvCommentOuput.setText("Number too high!");
            noofguess = noofguess + 1;
        } else{
            tvCommentOuput.setText("Number too low!");
            noofguess = noofguess + 1;
        } 


        tvCommentOuput.setText("Please enter a number between 1-20!");


        randtext = Integer.toString(n);

}
}

我想要做的是在if语句中围绕它，这样如果noofguess小于3，它将运行常规程序，但如果它高于3，那么它只会显示它们已经用尽猜测但是似乎不起作用。

Answer 1

解决方案是使用while循环包围部分代码

String randtext ="";
Random rand = new Random();
int noofguess = 0;
int n = rand.nextInt(20)+1; //generate random number
do{
int userguess =     Integer.parseInt(etCommentInput.getText().toString());
if (userguess < 1 || userguess > 20){
    tvCommentOuput.setText("Please enter a number between 1-20!");
}
else if (userguess == n){
    tvCommentOuput.setText("You Got It Right!");
    break;
} else if (userguess > n) {
    tvCommentOuput.setText("Number too high!");
    noofguess = noofguess + 1;
} else{
    tvCommentOuput.setText("Number too low!");
    noofguess = noofguess + 1;
} 


tvCommentOuput.setText("Please enter a number between 1-20!");


randtext = Integer.toString(n);
}
while(noofguess < 3);
// add if statement here to deal with the case where there are too many guesses

我如何制作它以便一旦超过阈值就会显示不同的消息？

1 个答案: