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Question

我遇到了這部分代碼的問題。該程序要求輸入一對夫婦（姓名，身份證，成績等），然後將結果打印回來。

我決定從本教程割捨，現在已嫌我的腦袋上一個衆所周知的牆 -

僞什麼，我想在這裏：

Ask user for grade between 9 and 12 
If input is less than 9 or greater than 12, return failed message and -return to loop- 
If input acceptable, continue to next question. 

當前代碼如下：

do { 
    System.out.print("Grade (9-12): "); 
    while (!keyboard.hasNextInt()) { 
     System.out.printf("message saying you're wrong"); 
     keyboard.next(); 
    } 
    userGrade = keyboard.nextInt(); 
} while (userGrade >= 9 || userGrade <= 12); 

Answer 1

的東西嘗試這樣的：

boolean correct = true; 
do { 
    System.out.print("Grade (9-12): "); 
    userGrade = keyboard.nextInt(); 
    if (userGrade < 9 || userGrade > 12) { 
     correct = false; 
     System.out.println("message saying you're wrong"); 
    } else { 
     correct = true; 
    } 
} while (!correct); 

Answer 2

我認爲這個問題是在邏輯...

變化

while (userGrade >= 9 || userGrade <= 12); 

到：

while (userGrade >= 9 && userGrade <= 12); 

的||接受高於等於9和低於等於12的任何東西。最後兩個條件使任何整數在條件中都成立。

Answer 3

您可以將您的任務分成較小的任務。例如創建的helper方法，其

• 將來自掃描儀讀取直到它找到整數，然後將返回

  public static int getInt(Scanner scanner, String errorMessage){ 
      while (!scanner.hasNextInt()) { 
       System.out.println(errorMessage); 
       scanner.nextLine(); 
      } 
      return scanner.nextInt(); 
  } 
  

• 或將檢查數量在範圍內（但也只是爲了可讀性）

  public static boolean isInRange(int x, int start, int end){ 
      return start <= x && x <= end; 
  } 
  

所以用這個方法，你的代碼可以像

Scanner scanner = new Scanner(System.in); 
int x; 

System.out.println("Please enter a number in range 9-12:"); 
do { 
    x = getInt(scanner, "I said number. Please try again: "); 
    if (!isInRange(x, 9, 12)) 
     System.out.println("I said number in range 9-12. Please try again: "); 
} while (!isInRange(x, 9, 12)); 

System.out.println("your number is: " + x);