Java 石头剪纸游戏不会运行

好的，大家好。这是我第一次在stackoverflow上，我在这里做了很多搜索，所以我想这将是一个获得一些答案的好地方。我是一个初学者程序员，目前在学校。我正在完成我的任务，制作一个石头、布、剪刀的游戏。到目前为止，我觉得我的代码很好，我编译了它，但它根本不会运行。看一看:

    import java.util.Scanner;
    public class RPSGame {


    public static void gameModeSelect ()
    {
    System.out.println("Welcome to Rock, Paper, Scissors 1.0 !\n Please select your game mode: ");
    System.out.println("1. Player vs. Computer\n 2. Player vs. Player ");
    }

    public static void winLoss ()
    {
    int P1 = 0, P2 = 0;

    if (P1 == 'P' && P2 == 'R'){
        System.out.println("Paper covers rock!\nPlayer one wins!");
    } else if (P1 == 'R' && P2 == 'P'){
        System.out.println("Paper covers rock!\nPlayer two wins!");
    }    
    if (P1 == 'R' && P2 == 'S'){
        System.out.println("Rock breaks scissors!\nPlayer one wins!");
    } else if (P1 == 'S' && P2 == 'R'){
        System.out.println("Rock breaks scissors!\nPlayer two wins!");
    }
    if (P1 == 'S' && P2 == 'P'){
        System.out.println("Scissor cuts paper!\nPlayer one wins!");
    } else if (P1 == 'P' && P2 == 'S'){
        System.out.println("Scissor cuts paper!\nPlayer two wins!");
    }
   }

    public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);  

    int P1;
        P1 = 0;
    int P2;
        P2 = 0;
    int modeSelect;
    modeSelect = keyboard.nextInt();

    gameModeSelect ();
    if (modeSelect == 1){
        System.out.println("Oops, this feature in currently unavailable. Play with a friend for now :-)");
    } else if (modeSelect == 2){

        System.out.println("Rules of the game:  R = Rock, P = Paper, S = Scissors\n Good luck! ");

        System.out.println("Player one: Enter your move");
        P1 = keyboard.nextInt();
        System.out.println("Player two: Enter your move");
        P2 = keyboard.nextInt();

    } else if (modeSelect > 2){
        winLoss();

    }

  }
}

它编译时没有错误，但没有运行，但是当我在“运行”部分输入任何内容时，会出现以下错误：

run:
S
Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Scanner.java:864)
    at java.util.Scanner.next(Scanner.java:1485)
    at java.util.Scanner.nextInt(Scanner.java:2117)
    at java.util.Scanner.nextInt(Scanner.java:2076)
    at RPSGame.main(RPSGame.java:41)
C:\Users\AVLG2\AppData\Local\NetBeans\Cache\8.2\executor-snippets\run.xml:53: Java returned: 1
BUILD FAILED (total time: 12 seconds)

---

我被困在这里了，我的作业明天就要交了。有什么建议吗？我们将不胜感激。顺便说一句，很高兴成为stackoverflow社区的一员

因为在

main

方法中调用了

modeSelect = keyboard.nextInt();

在打印任何东西之前。此调用将停止主线程，直到用户提供一些要读取的数据。如果数据类型为

int

控制流将把您移动到下一行，如果数据类型为

int

将抛出异常（就像您的情况一样，因为

R

无效

int

）

要让用户知道正在进行的操作，请打印说明程序正在等待其输入的消息：

System.out.print("Please provide game mode (<mode explanation here>): ");
modeSelect = keyboard.nextInt();

System.out.print（“请提供游戏模式（）：”；
modeSelect=keyboard.nextInt（）；

另外，当您向用户询问字符时，您不能使用

nextInt（）

，因为

R

不是

int

。改用

next（）

---

如果出现问题，请阅读：

---

如前所述，

扫描仪#nextInt（）

只能解析用户输入的整数。对于字符，它抛出一个异常

以下是您的代码，并对其进行了一些小的更改：

public class RPSGame {

public static void gameModeSelect() {
    System.out.println("Welcome to Rock, Paper, Scissors 1.0 !\n Please select your game mode: ");
    System.out.println("1. Player vs. Computer\n 2. Player vs. Player ");
}

public static void winLoss() {
    int P1 = 0, P2 = 0;

    if (P1 == 'P' && P2 == 'R') {
        System.out.println("Paper covers rock!\nPlayer one wins!");
    } else if (P1 == 'R' && P2 == 'P') {
        System.out.println("Paper covers rock!\nPlayer two wins!");
    }
    if (P1 == 'R' && P2 == 'S') {
        System.out.println("Rock breaks scissors!\nPlayer one wins!");
    } else if (P1 == 'S' && P2 == 'R') {
        System.out.println("Rock breaks scissors!\nPlayer two wins!");
    }
    if (P1 == 'S' && P2 == 'P') {
        System.out.println("Scissor cuts paper!\nPlayer one wins!");
    } else if (P1 == 'P' && P2 == 'S') {
        System.out.println("Scissor cuts paper!\nPlayer two wins!");
    }
}

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);

    String P1 = null;
    String P2 = null;
    int modeSelect;

    gameModeSelect();
    modeSelect = keyboard.nextInt();
    if (modeSelect == 1) {
        System.out.println("Oops, this feature in currently unavailable. Play with a friend for now :-)");
    } else if (modeSelect == 2) {

        System.out.println("Rules of the game:  R = Rock, P = Paper, S = Scissors\n Good luck! ");

        System.out.println("Player one: Enter your move");
        P1 = keyboard.next();
        System.out.println("Player two: Enter your move");
        P2 = keyboard.next();

    } else if (modeSelect > 2) {
        winLoss();

    }

}
}

缺少评估，但我想这取决于你

public static void gameModeSelect() {
    System.out.println("Welcome to Rock, Paper, Scissors 1.0 !\n Please select your game mode: ");
    System.out.println("1. Player vs. Computer\n 2. Player vs. Player ");
}

public static void winLoss(char P1, char P2) {

    if (P1 == 'P' && P2 == 'R') {
        System.out.println("Paper covers rock!\nPlayer one wins!");
    } else if (P1 == 'R' && P2 == 'P') {
        System.out.println("Paper covers rock!\nPlayer two wins!");
    }
    if (P1 == 'R' && P2 == 'S') {
        System.out.println("Rock breaks scissors!\nPlayer one wins!");
    } else if (P1 == 'S' && P2 == 'R') {
        System.out.println("Rock breaks scissors!\nPlayer two wins!");
    }
    if (P1 == 'S' && P2 == 'P') {
        System.out.println("Scissor cuts paper!\nPlayer one wins!");
    } else if (P1 == 'P' && P2 == 'S') {
        System.out.println("Scissor cuts paper!\nPlayer two wins!");
    }
}

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);

    char P1;
    char P2;
    int modeSelect;

    gameModeSelect();
    modeSelect = keyboard.nextInt();

    if (modeSelect == 1) {
        System.out.println("Oops, this feature in currently unavailable. Play with a friend for now :-)");
    } else if (modeSelect == 2) {

        System.out.println("Rules of the game:  R = Rock, P = Paper, S = Scissors\n Good luck! ");

        System.out.println("Player one: Enter your move");
        P1 = keyboard.next().charAt(0);
        System.out.println("Player two: Enter your move");
        P2 = keyboard.next().charAt(0);
        winLoss(P1, P2);

    }
}

您必须选择游戏模式：modeSelect=keyboard.nextInt（）

读取玩家输入：P1=键盘.next（）.charAt（0）；（&P2行）

评估游戏：使用适当的值调用winLoss函数

---

我不确定这个回复是否被允许，因为我读到了“只有问题和答案”，但谢谢！我最初试图使P1和P2字符，但无法使其工作。你建议的“keyboard.next（）.charAt（0）；”是我找不到的关键。再次感谢。我想我是从这里得到的！可能重复的