Java在「無效輸入」上循環切換case語句

Question

我正在尋找更新我的代碼來循環我的swtich語句，如果用戶輸入任何沒有定義的選項的話。我已經搜遍了各種各樣的搜索術語返回的許多頁面，並且已經接近，但迄今爲止還沒有運氣。 我在這裏的代碼應該得到任何想要刺穿它的人。

java.util.Scanner; 
//import java.lang.Character.*; 
//Thought this was needed to grab single char but its not 
public class caseloop { 
//main Method 
public static void main(String[] args) 
{ 
    Scanner input=new Scanner(System.in); //make so you can give input 
    boolean go = true; // for starting main outer loop 
    boolean run=true; // start inner loop 
    while (go==true) 
    { 
    while (run==true) 
    { 
     //Output 
     System.out.println("Enter option \n 1-Do this \n 2-Do this thing \n 3-Do this other thing"); 
     int option= input.nextInt(); //grab option number 

     switch(option) 
     { 
     /* 
      * This needs to loop and prompt user again if anything other than 1,2, or 3 is entered. 
      */ 
     case 1: 
      System.out.println("Option1"); 
      break; 
     case 2: 
      System.out.println("Option2"); 
      break; 
     case 3: 
      System.out.println("Option3"); 
      break; 
     /*case 4: 
      System.out.println("Option1"); 
      System.out.println("Option2"); 
      System.out.println("Option3"); 

      break; 
     * 
     * 
     * Case 4 was for debug 
     * 
     */ 
     default: 
      System.err.println("Invalid option selected"); 
      /* 
      * On input that is not defined with in the switch-case it will revert to "default" 
      * this fault staement needs to tell ther usere their option is not vaild and then 
      * prompt them to try it again to enter an option. I can not get it to reprompt. 
      * I have tried a while and an if loop both sorta worked but did not actually loop 
      * back to display again. I have been instucted that I am to not use a try catch statment 
      * unless of course that is the only viable option in whichcase I will use it anyways. 
      */ 

      //stupid default statement and its redundent built in "break;" 

     } 
     run=false; 
     } 


    /* 
    * Outer Loop to prompt user if they want to run the entire program again with new entries. 
    */ 
    if (run == false) 
    { 
    System.out.println("Would you like to run again? Y/N"); 
    char again = input.next().charAt(0); 
    again = Character.toUpperCase(again); //force all leters inputed to upper case, lower would work too if i change if conditions 
    if (again == 'Y') 
    { 
     run = true; 
    } 
    else if (again == 'N') 
    { 
     System.out.println("Goodbye."); 
     go=false; 
    } 
    else 
    { 
     System.err.println("Invalid entry. Try again."); 
    } 
    } 
    } 
} 
    //System.err.println("An error occured please try again"); 

} 

任何在這方面的援助將不勝感激。

Answer 1

您正在以非常奇怪的方式使用運行變量。由於您在循環結束時將運行設置爲false，因此循環將不會重複。如果你改變它，只有有效的選項設置run=false，輸入錯誤的選項會導致循環再次運行。

在switch聲明的末尾刪除run=false，則System.out.println("OptionX");

Answer 2

「運行=假」的部分是不是在正確的地方之後添加它，它執行任何答案（有效與否）。

您應該移動「run = false;」每個有效的情況下語句中，如：

case 1: 
     System.out.println("Option1"); 
     run=false; 
     break; 

順便說一句：「雖然（運行==真）」是多餘的，你可以寫「而（運行）」

Answer 3

的問題是，執行後語句默認情況下，它將布爾運行設置爲false，並因此出現循環。我們需要的是一種跳過此方法： -

run = false; 

而是直接轉到循環條件。一種解決方案是在默認情況下添加「繼續」語句： -

switch(option) 
    { 
    /* 
     * This needs to loop and prompt user again if anything other than 1,2, or 3 is entered. 
     */ 
    case 1: 
     System.out.println("Option1"); 
     break; 
    case 2: 
     System.out.println("Option2"); 
     break; 
    case 3: 
     System.out.println("Option3"); 
     break; 
    /*case 4: 
     System.out.println("Option1"); 
     System.out.println("Option2"); 
     System.out.println("Option3"); 

     break; 
    * 
    * 
    * Case 4 was for debug 
    * 
    */ 
    default: 
     System.err.println("Invalid option selected"); 
     continue; //this causes control to go back to loop condition 

    } 

Answer 4

您可以將標籤添加到循環中。這樣，當你有正確的選擇時，你將退出循環。

loop: switch(option) 
    { 
/* 
    * This needs to loop and prompt user again if anything other than 1,2, or 3 is entered. 
    */ 
case 1: 
    System.out.println("Option1"); 
    break loop; 
case 2: 
    System.out.println("Option2"); 
    break loop; 
case 3: 
    System.out.println("Option3"); 
    break loop; 
/*case 4: 
    System.out.println("Option1"); 
    System.out.println("Option2"); 
    System.out.println("Option3"); 

    break; 
* 
* 
* Case 4 was for debug 
* 
*/ 
default: 
    System.err.println("Invalid option selected"); 
    continue; //this causes control to go back to loop condition 

}