代碼執行沒有全部原因

Question

我正在做一個小程序，用戶選擇哪個選項想要做的。我使用開關做，並且一切正常，但現在我試圖增加選擇語言的能力，我試圖用數組來完成這個（我不知道如何正確解釋它，但是很容易請參閱代碼）。

//Each menu option 
static String[][] menuOptions = 
{ 
    { 
     "1. Check your grades", 
     "2. Check if you can examine for a driving license", 
     "3. Check if a number is odd or even", 
     "4. Check if a number is divisible by 7", 
     "5. Check if a triangle is equilater", 
     "6. Check who won", 
     "7. Check which number is bigger", 
     "8. Check if you can have a \"Carnet Jove\"", 
     "9. Convert numeric grades to letters", 
     "10. Exit Application" 
    }, 
    { 
     "1. Comprobar si has aprobado", 
     "2. Check if you can examine for a driving license", 
     "3. Check if a number is odd or even", 
     "4. Check if a number is divisible by 7", 
     "5. Check if a triangle is equilater", 
     "6. Check who won", 
     "7. Check which number is bigger", 
     "8. Check if you can have a \"Carnet Jove\"", 
     "9. Convert numeric grades to letters", 
     "10. Exit Application" 
    } 
}; 
//End of Options 

static Scanner Scan = new Scanner(System.in); 

static boolean correctInput = false; 
static int language; 
static int selection; 

public static void main(String[] args) 
{ 
    System.out.println("\t1. English\t2. Español\n"); 
    System.out.print("Select a Language:\tSeleccione un Idioma:\t"); 
    language = Scan.nextInt(); 
    System.out.print(""); 
    while (correctInput != true) 
    { 
     menu(language); 
     try //Comprobamos que el usuario haya introducido un numero 
     { 
      selection = Integer.parseInt(Scan.nextLine()); 
     } catch (NumberFormatException ex) //En caso de error lo gestionamos 
     { 
      //No hacemos nada 
     } 

     switch (selection) 
     { 
      case 1: 
       correctInput = true; 
       checkGrades(); 
       break; 
      case 2: 
       correctInput = true; 
       chechDrivingLicense(); 
       break; 
      case 3: 
       correctInput = true; 
       checkOddNum(); 
       break; 
      case 4: 
       correctInput = true; 
       checkDivBy7(); 
       break; 
      case 5: 
       correctInput = true; 
       checkEquilater(); 
       break; 
      case 6: 
       correctInput = true; 
       checkWinner(); 
       break; 
      case 7: 
       correctInput = true; 
       checkBigger(); 
       break; 
      case 8: 
       correctInput = true; 
       checkCarnetJove(); 
       break; 
      case 9: 
       correctInput = true; 
       convertNumGradeToLetter(); 
       break; 
      case 10: 
       correctInput = true; 
       break; 
      default: 
       System.out.println("\n\n\n\nInput not valid. Please enter" 
         + " a valid number\n"); 
     } 
    } 

} 

private static void menu(int language) 
{ 
    System.out.println("\n"); 
    int sel = 12; 
    for (String s : menuOptions[language - 1]) 
    { 
     System.out.println("\t" + s); 
    } 
    System.out.print("\nSelect an option: \t"); 

} 

而不是顯示所選菜單的（對於現在只有翻譯一個選項，但也足以檢查是否正常工作與否），什麼情況是，一旦顯示菜單，它會自動選擇一個選項（這總是無效的）並且觸發菜單重複1次。這不是一個大麻煩，但我想解決它。

這裏有一個電流輸出的樣子：

Answer 1

language = Scan.nextInt(); 

這讀取一個整數，而不是後面的換行符。

selection = Integer.parseInt(Scan.nextLine()); 

第一次到達時，有一個換行符正在等待閱讀。 nextLine()立即返回一個空字符串。

catch (NumberFormatException ex) //En caso de error lo gestionamos 
{ 
    //No hacemos nada 
} 

如果您沒有吞下產生的異常，您可能會看到這一點。不要無所事事地處理異常！至少請致電ex.printStackTrace()查看錯誤消息。更好的是，請用戶再試一次。

要修復您的程序，請避免混合使用nextInt()和nextLine()。總是使用nextLine()更好。閱讀language就像你做selection一樣，你會變得更好。

language = Integer.parseInt(Scan.nextLine());