如何使用try catch java？

Question

對於我的任務之一，我們應該嘗試捕獲一個InputMismatchException並告訴用戶輸入一個給定的選項（在這種情況下，它可以是1,2,3,4而沒有其他）。出於某種原因，每當我嘗試創建try catch塊時，它都會讓用戶輸入兩次輸入，並且仍然會使程序崩潰。這裏是我有：

do { 
     Scanner menuOptions = new Scanner(System.in); 
     System.out.println("1. Get another card"); 
     System.out.println("2. Hold hand"); 
     System.out.println("3. Print statistics"); 
     System.out.println("4. Exit"+ "\n"); 
     System.out.println("Choose an Option: "); 
     int selectedOption = menuOptions.nextInt(); 

     try{ 
     selectedOption = menuOptions.nextInt(); 
     } 

     catch(InputMismatchException e){ 
     System.out.println("Invalid input! Please select between options 1,2,3,4") 
     } 


     if (selectedOption == 1) { 
     //proceeds to do what is instructed if user input is 1} 
     else if (selectedOption ==2{ 
     //proceeds to do what is instructed if user input is 2} 
    else if (selectedOption == 3) { 
     //proceeds to do what is instructed if user input is 3} 
    else if (selectedOption ==4{ 
     //proceeds to do what is instructed if user input is 4} 

任何想法，我怎麼能使這項工作？

Answer 1

「進入輸入兩次」

因爲你有兩行要求的數字：

menuOptions.nextInt(); 

Answer 2

這是因爲你叫menuOptions.nextInt()兩次。初始化完成後，第二次在try catch塊中。我假設你有你想要運行，直至用戶輸入有效輸入代碼，是1，2，3或4

試試這個：

int selectedOption; 
while(true) { 
    try { 
     selectedOption = menuOptions.nextInt(); 
     break; 
    } catch(InputMismatchException ime) { 
     System.out.println("Invalid input! Please select between options 1,2,3,4"); 
     menuOptions = new Scanner(System.in); 
    } 
}