Question

我在为自己做一个小游戏时遇到一些问题。我遇到问题的代码是用户输入超时。例如：游戏将在屏幕上显示一个字母。用户必须在一定时间内输入该字母;如果他们不这样做，就会受到伤害。如果有人可以解释我可以用什么来解决这个问题，那就太好了。

package weapons;

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


class wepons {
    public static void main(String args[]){

        int weapondamage =0 , health = 0 , hits = 0, potion = 0, beast = 0, beastdmg = 0, letter = 0;
        String weapon;
        ;

        Scanner in = new Scanner (System.in);   

        System.out.println ("please pick a wepon that will kill the boss \n");
        weapon = in.nextLine().toLowerCase();
        if (weapon.equals("sword")){
          System.out.println("you have picked the sword");
          weapondamage = (int) ((Math.random()*100));
          health = weapondamage*3;
        }


        System.out.println("Sword damage is " + weapondamage);
        System.out.println("Boss health is " + health);
        System.out.println("Enter the amount of hits it will take to kill the boss.");
        hits = in.nextInt();
        if(hits == health/weapondamage) {
        System.out.println("you have killed the boss.The boss had droped a potion.\nYou pick up the potion and drink it!\nYour health has now gone up to 350hp");
        }

        else {

            System.out.print("You have failed to kill the boss");



        }
        potion = (int)350;
        beastdmg = (int) ((Math.random()*60));
        System.out.println("By killing the boss you have awoken the beast!");
        System.out.println("The beast nocks you onto your back");
        System.out.println("The beast will hit you for"+beastdmg+"damage");
        System.out.println("You can block the beast damage by hitting the write key");
        Random r = new Random();
        int c = r.nextInt(26) + (byte)'a';
        System.out.println((char)c);
        letter = in.next().charAt(0);
        if(  letter == ((char)c)){
            System.out.println("You blocked the beast damage");




        }
        else {

            System.out.print("You took damage");



        }
        }




        }  // main

Answer 1

有多种方法可以做到这一点，我正在以最简单的方式回答使用OP的风格，所以测量用户按下字母的时间。这是一个功能正常的代码段：

    System.out.println("You can block the beast damage by hitting the write key");
    long startTime = System.currentTimeMillis();
    Random r = new Random();
    int c = r.nextInt(26) + (byte) 'a';
    System.out.println((char) c);
    char letter = in.next().charAt(0);
    long stopTime = System.currentTimeMillis();
    long elapsedTime = stopTime - startTime;
    if (letter == ((char) c) && elapsedTime <=5000) {
        System.out.println("You blocked the beast damage");

    } else {

        System.out.print("You took damage");

    }

希望它有所帮助。

Answer 2

你需要看一下Threads。

使用线程，您可以一次执行多个进程，而不是逐行执行线性编码。

例如，您可以说：

while(health > 0){
   Thread.sleep(5000); //parameter for sleep is milliseconds, so this is 5 seconds
   if(weapon == null){
      health -= damage; 
   }else{ break; }
}

这个问题是你可能需要创建一个单独的类来扩展Thread以调用损失...如果定时方面是必要的，我会建议读取线程。不要从字面上理解这个代码，这只是一个基本的大纲......根据我对线程的经验，你需要创建一个返回武器值的方法，因为你希望你的线程检查用户是否输入了值对于武器。

您可能还想参考这个线程教程：

https://docs.oracle.com/javase/tutorial/essential/concurrency/simple.html

用户输入的时间限制

2 个答案: