Question

我问用户他们是否想玩游戏..

    System.out.println("Would you like to play"); 
     read in yes or no value 
        if yes (display text) 
        else (display other text)

我之前在程序中使用过扫描仪，我只需要再次使用它。我是否需要使用新变量声明另一个新的变量？

Answer 1

你可以做点什么：

    String yesOrNo = System.in.readLine();
    String textToDisplay = (yesOrNo.equals("Yes")) ? "text to display when yesOrNo equals yes" : "text to display when YesOrNo equals no";

Answer 2

使用此方法：

scanner.nextBoolean()

那么它就是：

System.out.println("Would you like to play?");
Scanner scanner = new Scanner(System.in);

if(scanner.nextBoolean()==true) {
    System.out.println("This will be fun");
} else {
    System.out.println("Maybe next time");
}

修改

System.out.println("Would you like to play?");
Scanner scanner = new Scanner(System.in);
String val = scanner.next();
if(val.equalsIgnoreCase("y")||val.equalsIgnoreCase("yes")) {
    System.out.println("This will be fun");
} else if(val.equalsIgnoreCase("n")||val.equalsIgnoreCase("no")) {
    System.out.println("Maybe next time");
} else { 
System.out.println("Invalid character");

Answer 3

为了简单起见，您可以使用以下内容：

import java.util.Scanner;

public class ScannerExample {

  public static void main(String[] args) {

    System.out.println("Would you like to play: 'y' or 'yes' to accept; 'n' or 'no' to reject; 'q' to quit:");
    Scanner scanner = new Scanner(System.in);    
    String token = "";

    while(scanner.hasNextLine())
    {
       token = scanner.nextLine().trim();

       if(token.equalsIgnoreCase("q")) System.exit(0);

       if(token.equalsIgnoreCase("y")||token.equalsIgnoreCase("yes")) 
       {
           System.out.println("Thanks for your interest!");
           System.exit(0);
       }
       else if (token.equalsIgnoreCase("n")||token.equalsIgnoreCase("no"))
       {
           System.out.println("That's a pity!");
           System.exit(0);
       }
       else
       {
           System.out.println("Oops, not a valid input!");
       }
    }
  }
}

Answer 4

我只是偶然发现了这个问题，而且Chad M的答案是正确的，但是 - 就像上面提到的一些评论者 - 有一些错误。所以这是我没有错误的实现（这只是将用户输入保存到字符串变量的问题）：

Scanner scanner = new Scanner(System.in);
String userInput = scanner.next();

if(userInput.equalsIgnoreCase("y") || userInput.equalsIgnoreCase("yes")) {
    // y or yes
} else {
    // other character
}

Answer 5

根据dragon66的推荐，我正在实现与用户输入相比预定义输入的字符串比较。

System.out.println("Would you like to play? Or press Q to exit"):
    Scanner scanner = new Scanner(System.in);
    String affirmative = "yes";
    String affirmative2 = "y";
    String negative = "no";
    String negative2 = "n";
    String quit = "q";

if(scanner.next().equalsIgnoreCase(affirmative)||scanner.next().equalsIgnoreCase(affirmative2)) {
    System.out.println("This will be fun");
} else if(scanner.next().equalsIgnoreCase(negative)||scanner.next().equalsIgnoreCase(negative2)) {
    System.out.println("Maybe next time");
} else if(scanner.next().equalsIgnoreCase(quit)){ 
    System.out.println("Good bye");
    System.exit();
} else {
System.out.println("Invalid character");
}

从终端读入yes / no boolean并生成带有输入值的if else

5 个答案: