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Question

我的这部分代码有问题。该程序要求输入几个输入（姓名、ID、等级等），然后将结果打印回来。

我决定脱离教程，现在一直在用众所周知的墙壁砸我的头 -

我想要的伪代码：

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. 

当前代码如下：

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

试试这样的：

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

我认为问题出在逻辑上……

改变

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

到：

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

||接受任何大于等于 9 和小于等于 12 的值。这两个条件最终使得任何整数在条件下都为真。

Answer 3

您可以将任务拆分为较小的任务。例如创建

的辅助方法

• 将从 Scanner 读取直到找到整数，然后将其返回

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

• 或将检查数字是否在范围内（但这只是为了便于阅读）

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

因此，使用这种方法，您的代码可以看起来像

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);