石头、纸、剪刀游戏 Java
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原文地址: http://stackoverflow.com/questions/19204872/
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Rock, Paper, Scissors Game Java
提问by
I'm new to programming and I'm trying to write a very simple Rock, Paper, Scissors game in Java. It will compile and run fine, but I am looking to say something like "Invalid move. Try again." or something along those lines for when the user (personPlay) does not enter a correct character (r, p, or s). What would be the best way to do so? For example, if you enter a "q", it should print "Invalid move." Thank you so much in advance!
我是编程新手，我正在尝试用 Java 编写一个非常简单的石头剪刀布游戏。它将编译并运行良好，但我想说“无效的移动。再试一次”。或者当用户 (personPlay) 没有输入正确的字符（r、p 或 s）时的类似内容。这样做的最佳方法是什么？例如，如果您输入“q”，它应该打印“Invalid move”。非常感谢您！
// *************
// Rock.java
// *************
import java.util.Scanner;
import java.util.Random;
public class Rock
{
public static void main(String[] args)
{
String personPlay; //User's play -- "R", "P", or "S"
String computerPlay = ""; //Computer's play -- "R", "P", or "S"
int computerInt; //Randomly generated number used to determine
//computer's play
String response;
Scanner scan = new Scanner(System.in);
Random generator = new Random();
System.out.println("Hey, let's play Rock, Paper, Scissors!\n" +
"Please enter a move.\n" + "Rock = R, Paper" +
"= P, and Scissors = S.");
System.out.println();
//Generate computer's play (0,1,2)
computerInt = generator.nextInt(3)+1;
//Translate computer's randomly generated play to
//string using if //statements
if (computerInt == 1)
computerPlay = "R";
else if (computerInt == 2)
computerPlay = "P";
else if (computerInt == 3)
computerPlay = "S";
//Get player's play from input-- note that this is
// stored as a string
System.out.println("Enter your play: ");
personPlay = scan.next();
//Make player's play uppercase for ease of comparison
personPlay = personPlay.toUpperCase();
//Print computer's play
System.out.println("Computer play is: " + computerPlay);
//See who won. Use nested ifs
if (personPlay.equals(computerPlay))
System.out.println("It's a tie!");
else if (personPlay.equals("R"))
if (computerPlay.equals("S"))
System.out.println("Rock crushes scissors. You win!!");
else if (computerPlay.equals("P"))
System.out.println("Paper eats rock. You lose!!");
else if (personPlay.equals("P"))
if (computerPlay.equals("S"))
System.out.println("Scissor cuts paper. You lose!!");
else if (computerPlay.equals("R"))
System.out.println("Paper eats rock. You win!!");
else if (personPlay.equals("S"))
if (computerPlay.equals("P"))
System.out.println("Scissor cuts paper. You win!!");
else if (computerPlay.equals("R"))
System.out.println("Rock breaks scissors. You lose!!");
else
System.out.println("Invalid user input.");
}
}
}
回答by Serge Seredenko
You could insert something like this:
你可以插入这样的东西：
personPlay = "B";
while (!personPlay.equals("R") && !personPlay.equals("P") && !personPlay.equals("S")) {
//Get player's play from input-- note that this is
// stored as a string
System.out.println("Enter your play: ");
personPlay = scan.next();
//Make player's play uppercase for ease of comparison
personPlay = personPlay.toUpperCase();
if (!personPlay.equals("R") && !personPlay.equals("P") && !personPlay.equals("S"))
System.out.println("Invalid move. Try again.");
}
回答by dkatzel
I would recommend making Rock, Paper and Scissors objects. The objects would have the logic of both translating to/from Strings and also "knowing" what beats what. The Java enum is perfect for this.
我建议制作石头、纸和剪刀物体。对象将具有与字符串相互转换的逻辑，并且还具有“知道”什么胜过什么的逻辑。Java 枚举非常适合于此。
public enum Type{
ROCK, PAPER, SCISSOR;
public static Type parseType(String value){
//if /else logic here to return either ROCK, PAPER or SCISSOR
//if value is not either, you can return null
}
}
The parseTypemethod can return nullif the String is not a valid type. And you code can check if the value is null and if so, print "invalid try again" and loop back to re-read the Scanner.
如果 String 不是有效类型，则该parseType方法可以返回null。并且您的代码可以检查该值是否为空，如果是，则打印“无效重试”并循环返回以重新读取扫描仪。
Type person=null;
while(person==null){
System.out.println("Enter your play: ");
person= Type.parseType(scan.next());
if(person ==null){
System.out.println("invalid try again");
}
}
Furthermore, your type enum can determine what beats what by having each Typeobject know:
此外，您的类型枚举可以通过让每个Type对象知道什么来确定什么胜过什么：
public enum Type{
//...
//each type will implement this method differently
public abstract boolean beats(Type other);
}
each type will implement this method differently to see what beats what:
每种类型都会以不同的方式实现这个方法，看看什么胜过什么：
ROCK{
@Override
public boolean beats(Type other){
return other == SCISSOR;
}
}
...
Then in your code
然后在你的代码中
Type person, computer;
if (person.equals(computer))
System.out.println("It's a tie!");
}else if(person.beats(computer)){
System.out.println(person+ " beats " + computer + "You win!!");
}else{
System.out.println(computer + " beats " + person+ "You lose!!");
}
回答by musical_coder
Before we try to solve the invalid character problem, the lack of curly braces around the ifand else ifstatements is wreaking havoc on your program's logic. Change it to this:
在我们尝试解决无效字符问题之前，ifandelse if语句周围缺少大括号对您的程序逻辑造成了严重破坏。改成这样：
if (personPlay.equals(computerPlay)) {
System.out.println("It's a tie!");
}
else if (personPlay.equals("R")) {
if (computerPlay.equals("S"))
System.out.println("Rock crushes scissors. You win!!");
else if (computerPlay.equals("P"))
System.out.println("Paper eats rock. You lose!!");
}
else if (personPlay.equals("P")) {
if (computerPlay.equals("S"))
System.out.println("Scissor cuts paper. You lose!!");
else if (computerPlay.equals("R"))
System.out.println("Paper eats rock. You win!!");
}
else if (personPlay.equals("S")) {
if (computerPlay.equals("P"))
System.out.println("Scissor cuts paper. You win!!");
else if (computerPlay.equals("R"))
System.out.println("Rock breaks scissors. You lose!!");
}
else
System.out.println("Invalid user input.");
Much clearer! It's now actually a piece of cake to catch the bad characters. You need to move the elsestatement to somewhere that will catch the errors beforeyou attempt to process anything else. So change everything to:
清晰多了！现在抓住坏角色实际上是小菜一碟。在尝试处理其他任何事情之前else，您需要将语句移动到可以捕获错误的地方。因此，将所有内容更改为：
if( /* insert your check for bad characters here */ ) {
System.out.println("Invalid user input.");
}
else if (personPlay.equals(computerPlay)) {
System.out.println("It's a tie!");
}
else if (personPlay.equals("R")) {
if (computerPlay.equals("S"))
System.out.println("Rock crushes scissors. You win!!");
else if (computerPlay.equals("P"))
System.out.println("Paper eats rock. You lose!!");
}
else if (personPlay.equals("P")) {
if (computerPlay.equals("S"))
System.out.println("Scissor cuts paper. You lose!!");
else if (computerPlay.equals("R"))
System.out.println("Paper eats rock. You win!!");
}
else if (personPlay.equals("S")) {
if (computerPlay.equals("P"))
System.out.println("Scissor cuts paper. You win!!");
else if (computerPlay.equals("R"))
System.out.println("Rock breaks scissors. You lose!!");
}
回答by Srihari
Why not check for what the user entered and then ask the user to enter correct input again?
为什么不检查用户输入的内容，然后要求用户再次输入正确的输入？
eg:
例如：
//Get player's play from input-- note that this is
// stored as a string
System.out.println("Enter your play: ");
response = scan.next();
if(response=="R"||response=="P"||response=="S"){
personPlay = response;
}else{
System.out.println("Invaild Input")
}
for the other modifications, please check my total code at pastebin
对于其他修改，请在pastebin检查我的总代码
回答by Peter Kim
int w =0 , l =0, d=0, i=0;
Scanner sc = new Scanner(System.in);
// try tentimes
while (i<10) {
System.out.println("scissor(1) ,Rock(2),Paper(3) ");
int n = sc.nextInt();
int m =(int)(Math.random()*3+1);
if(n==m){
System.out.println("Com:"+m +"so>>> " + "draw");
d++;
}else if ((n-1)%3==(m%3)){
w++;
System.out.println("Com:"+m +"so>>> " +"win");
}
else if(n >=4 )
{
System.out.println("pleas enter correct number)");
}
else {
System.out.println("Com:"+m +"so>>> " +"lose");
l++;
}
i++;
