2

我正在尝试用java制作一种打字冒险游戏，但是我需要一个至少类似于标题中的命令，这是代码

import java.util.Scanner;

public class MyFirstGameInJava {

public static void main(String[] args) {

System.out.println("Greetings, Enter your name and you may start your quest!");
Scanner Username = new Scanner(System.in);
String name = Username.nextLine();
System.out.println("Greetings " + name );
System.out.println("Welcome to an Unnamed Typing Advanture");
System.out.println("You find yourself on an island with very few trees, you can either hit a tree, or walk along");

String sc = Username.nextLine();

switch(sc){

case "Hit tree":
System.out.println("A coconut falls from the tree");
System.out.println("You can either eat the coconut or throw it");
break;
case "Walk":
System.out.println("You walk for a mile and find a village");
System.out.println("The village appears empty, you can either scream to see if anybody is there, or you can keep walking");
break;
default :
System.out.println("Nothing happens...");
}   

String sc1 = Username.nextLine();


switch(sc1){

case "Eat coconut":
System.out.println("You ate the coconut and got poisoned");
System.out.println("You died...");
break;
case "Throw coconut":
System.out.println("By throwing the coconut, you awaken a tiger and he eats you");
System.out.println("You are dead");
break;
case "Scream":
System.out.println("As soon as you scream, a man shoots you down from a window from one of the houses");
System.out.println("You died...");
break;
case "Walk":
System.out.println("You walked through the village, and you find a boat and leave the island");
System.out.println("You win! Updates coming soon!");
break;
default:
System.out.print("Nothing happend");



}

}

}

每当用户键入非必需的内容时，就会发生默认情况，但我需要它返回到循环的开头，因此用户可以键入其他情况之一。

4

5 回答 5

7

您可以使用该continue语句继续下一次迭代。

也就是说，我在您的示例代码中没有看到循环。for您可以使用,while或循环do/while。do/while循环至少执行一次——这通常是您在向用户提问时想要执行的操作。

这个分支语句的 Java 教程提供了这个循环中的continue语句示例。for

   for (int i = 0; i < max; i++) {
        // interested only in p's
        if (searchMe.charAt(i) != 'p')
            continue;

        // process p's
        numPs++;
    }
于 2013-07-15T18:08:33.573 回答
3

continue;例如与无条件循环一起使用

while(true){/* your code*/}
于 2013-07-15T18:08:48.027 回答
0

您的问题做出了不正确的假设。switch()语句不是循环。它是一堆这样的 if/else-if 语句的简写：

if(sc1.equals("Eat coconut")) {
    System.out.println("You ate the coconut and got poisoned");
    System.out.println("You died...");
}
else if(sc1.equals("Throw coconut")) {
//and so on

为了做你想做的事，你需要一个实际的while()或for()循环。（这些链接将带您到解释这两种结构的 Java 教程。）

于 2013-07-15T18:10:17.753 回答
0

不那么详细，更直截了当。我已经解释了简单的答案。

for(int i = 0; i <= 10; i++){
        if(i == 4){
            continue;
        }
        //Should skip 4 and print out 1,2,3,5,6,7,8,9,10
        System.out.println(i);
    }
于 2017-01-19T14:22:13.343 回答
0

我的 CS 课也遇到了同样的问题：这就是我所做的。首先，您需要使用循环。我使用了“while”循环。

假设你在对话中的某个时刻有人向其他人询问某事（就像武术大师向玩家询问某事），而答案选择是“y”或“n”。但是玩家输入“m”。这会产生错误。要修复它，您需要制作一些东西（循环）来不断检查用户是否输入了有效的响应，并相应地继续程序。现在代码...

//Ask if they want to see the rules.    
System.out.println("Welcome Player One");  
System.out.println("Would you like to read the rules of the gem?");  
System.out.println("[y]Yes, please / [n]I know the rules."); //So you made it clear that the choices are 'y' or 'n' (yes or no).

//Decide if they hit yes or no.  
char Response; //Variable to hold user response.  
Scanner Console = new Scanner(System.in); //I don't know why, exactly, but you need this.  
Response = Console.next().charAt(0); //Set response to the user input. A input field will appear.  
while(true){  //Making the parameter 'true' makes the flow of the loop depend on the parameters in the if statements (I think).  
  if(Response == 'y' || Response == 'Y'){  
   System.out.println("Here are the rules:");  
   break; //break will make the flow exit the while loop (so you can continue adding the rest of your code).  
  }  
  else if(Response == 'n' || Response == 'N'){  
   System.out.println("Then let us continue:");  
   break;  
  }  
  else{  
   System.out.println("Would you like to read the rules of the gem?"  
   System.out.println("You have to enter (y)yes or (n)no...");  
   Response = Console.next().charAt(0); //Just like before, input field will  appear to give them the chance to change their response to a valid one.  
   continue; //This has to be 'continue' so that the loop will continue to "check" for the value that needs to be input by the user.  
  }  
}

希望这会有所帮助，干杯。

于 2015-09-02T21:33:22.160 回答