Question

好的，嗨，每个人。这是我第一次访问stackoverflow，我在这里进行了很多搜索，所以我认为这是一个获得答案的好地方。我是目前在学校的初学者。我正在完成创建摇滚，纸张，剪刀游戏的任务。到目前为止，我觉得我的代码很好，我编译它但它根本不会运行。看看：

    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，我收到此错误：

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，我的任务明天到期。有什么建议？非常感谢。顺便说一句，很高兴成为stackoverflow社区的一员!!

Answer 1

您没有看到任何内容，因为您调用了$pdo->lastInsertId();方法

main

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modeSelect = keyboard.nextInt();，则控制流将移动到下一行，如果不是int将抛出异常（就像您的情况int无效{ {1}}）。

让用户知道打印消息的内容，这解释了程序正在等待他的输入：

此外，当您要求用户输入字符时，您无法使用int，因为System.out.print("Please provide game mode (<mode explanation here>): "); modeSelect = keyboard.nextInt();不是nextInt()。请改用R。

如有问题，请阅读：

How do I compare strings in Java?
Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo() methods
What is a stack trace, and how can I use it to debug my application errors?

Answer 2

boolean x;
for(int i=0;i<=holdN.length;i++){
    ...
}
return x;

您必须选择游戏模式：modeSelect = keyboard.nextInt（）;
读取玩家输入：P1 = keyboard.next（）。charAt（0）; （＆amp; P2行）
评估游戏：使用适当的值调用winLoss函数

Answer 3

如前所述，Scanner#nextInt()只能解析用户输入的整数。对于角色，它会引发异常。

此处您的代码稍作修改，以使其有效：

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。

摇滚纸剪刀游戏不会运行

3 个答案: