Java数字猜谜游戏

时间:2015-10-14 04:45:31

标签: java numbers

我正在进行一个猜数字游戏,我在设置循环方面遇到了问题,让玩家可以再次玩游戏。我还想知道是否有办法在循环中包含提示选项。

我已经编程了2个月。

package carlosnumber;

import java.util.Random;
import java.util.Scanner;

public class CarlosNumber {

    public static void main(String[] args) {
        Scanner console = new Scanner(System.in);
        String name = "start";
        int number = 0;
        int numTries = 0;
        int secretNumber;
        String wantHints, playagain;
        int guess = -1, numGuess, wonGames, quitGames, playedGames;
        boolean end = false;
        boolean hint = false;
        boolean play = true;
        String win = null;

        System.out.println("Welcome the the Guessing game");

        while (play == true) {
            do {
                Random secretNumberGenerator = new Random();
                secretNumber = 56;

                if (name.equals("start")) {
                    System.out.println("To Start playing, please enter your name");
                    name = console.nextLine();

                    System.out.println("The system will randomly select a number between 1 and 100"
                            + "\nYour goal is to guess the number. At each turn enter"
                            + " in your guess, and I will tell you when your guess is correct");

                }
                System.out.println("Please Enter a number");
                number = console.nextInt();
                numTries++;

                if (number < secretNumber) {
                    System.out.println("Your guess is low");
                }
                if (number > secretNumber) {
                    System.out.println("Your guess is high");
                }
                if (number == secretNumber) {
                    System.out.println("YOU WIN!!" + "It took you " + numTries + " Tries");
                    end = true;
                }

            } while (end == false);

            System.out.println("Do you wish to play again? Y/N");
            playagain = console.nextLine();

            if (playagain == "Yes") {
                play = true;
            } else {
                play = false;
            }

        }
    }
}

0 个答案:

没有答案