在Java中提示用戶是或否輸入

Question

如何提示用戶循環代碼是循環否退出和錯誤輸入打印錯誤輸入並返回語句即。 「你想輸入另一個名字：」

import java.util.Scanner; 

public class loop { 
public static void main(String[] args){ 
    Scanner kbd = new Scanner (System.in); 
    String decision; 
    boolean yn; 
    while(true){ 

     System.out.println("please enter your name"); 
     String name = kbd.nextLine(); 

     System.out.println("you entered the name" + name); 

     System.out.println("enter another name : yes or no"); 
     decision = kbd.nextLine(); 

     switch(decision){ 
     case "yes": 
      yn = false; 
      break; 
     case "no": 
      yn = true; 
      break; 
     default : 
      System.out.println("please enter again "); 
      return default; 
     } 
    } 
    } 
} 

Answer 1

1. 如果不使用Java 7，你不能使用開關串

2. 變化while (true)與while (yn)所以它會停止時他鍵入「否」，並將boolean yn;更改爲boolean yn = true; 並更改案例中的規則。

   case "yes": 
       yn = false; 
       break; 
   case "no": 
       yn = true; 
       break; 
   

   yn = true;如果 「是」;

   yn = false; if「no」;

   你可以改變內部的條件，用while (!yn)，但更直觀的讓yntrue如果是的話;如果沒有，則爲false。

3. return default;沒有多大意義，如果你想要讓錯誤的情況下，用戶重複好..你應該做一個新的while (true)重複，直到他寫道正確的。我會寫另一種方法。

這是你如何能做到這一點

Scanner kbd = new Scanner (System.in); 

String decision; 

boolean yn = true; 
while(yn) 
{ 
    System.out.println("please enter your name"); 
    String name = kbd.nextLine(); 

    System.out.println("you entered the name" + name); 

    System.out.println("enter another name : yes or no"); 
    decision = kbd.nextLine(); 


    switch(decision) 
    { 
     case "yes": 
      yn = true; 
      break; 

     case "no": 
      yn = false; 
      break; 

     default: 
      System.out.println("please enter again "); 
      boolean repeat = true; 

      while (repeat) 
      { 
       System.out.println("enter another name : yes or no"); 
       decision = kbd.nextLine(); 

       switch (decision) 
       { 
        case "yes": 
         yn = repeat = true; 
         break; 

        case "no": 
         yn = repeat = false; 
         break; 
       } 
      } 
      break; 
    } 
} 

是的，它會重複decision代碼，但它是如何創建的，我認爲這是應該做的唯一途徑。

Answer 2

是的，但你想要while(yn)，不while(true)，這繼續下去。 break只從switch聲明中突破。

---

好吧，試試這個：

import java.util.Scanner; 

public static void main(String args[]){ 
    Scanner s = new Scanner(System.in); 

    boolean checking = true, valid = true; 
    String[] names = new String[50]; 
    int i = 0; 

    while(checking){ 
    System.out.println("Enter name..."); 
    me = s.nextLine(); 
    System.out.println("You entered " + me + "."); 
    while(valid){ 
     System.out.println("Enter another? y/n"); 
     you = s.nextLine(); 
     if(you.equals("n")){ 
     valid = false; 
     checking = false; 
     }else if you.equals("y")){ 
     names[i] = you; 
     i++; 
     valid = false; 
     }else{ 
     System.out.println("Sorry, try again (y/n)..."); 
     } 
    } 
    } 
} 

Answer 3

boolean input = true; 
while(input){ 
//ask for name 
//print message saying you entered xyz 
//ask if user wants to enter other name 
//if user enters no 
//set input = false; 
//else continue 
} 

嘗試這樣做

Answer 4

而是再突破，雖然（真），你需要說while(!yn)爲您YN設置爲false，當用戶輸入肯定的。

Answer 5

我剛剛創建了一個小班YesNoCommandPrompt是做到了這一點：

private String prompt; 
private Supplier<T> yesAction; 
private Supplier<T> noAction; 

public YesNoCommandPrompt(String prompt, Supplier<T> yesAction, Supplier<T> noAction) { 
    this.prompt = prompt; 
    this.yesAction = yesAction; 
    this.noAction = noAction; 
} 

// example usage 
public static void main(String[] args) { 
    YesNoCommandPrompt<String> prompt = new YesNoCommandPrompt<>(
      "Choose", 
      () -> {return "yes";}, 
      () -> {return "no";}); 

    System.out.println(prompt.run()); 
} 

public T run() { 
    final String yesOption = "y"; 
    final String noOption = "n"; 
    try (Scanner scanner = new Scanner(System.in)) { 
     while (true) { 
      System.out.print(prompt + " [" + yesOption + "/" + noOption + "]: "); 
      String option = scanner.next(); 
      if (yesOption.equalsIgnoreCase(option)){ 
       return yesAction.get(); 
      } 
      else if (noOption.equalsIgnoreCase(option)){ 
       return noAction.get(); 
      } 
     } 
    } 
}