java - 如何使用 InputMismatchException 捕获字符串？ ( java )

Question

我是一名新手 java 程序员，需要调整此代码，以便它捕获两个字符串而不是变量。

这是我们应该使用的原始代码:

import java.util.Scanner;
import java.util.InputMismatchException;

public class Part4 {

    public static void main(String[] args) {
        int userNum = 0;
        Scanner screen = new Scanner(System.in);
        boolean inputOK = false;
        String dump = null;
        while (!inputOK) {
            System.out.print("Enter a number: ");
            try {
                userNum = screen.nextInt();
                dump = screen.nextLine();
                inputOK = true;
            } catch (InputMismatchException e) {
                dump = screen.nextLine();
                System.out.println("\"" + dump + "\" is not a legal integer, " +
                        "please try again!");
            } // end try-catch block
        } // end input loop
        screen.close();
        userNum = userNum + 20;
        System.out.println("Your number plus 20 is " + userNum);
    }
} 

这是我失败的尝试:

import java.util.Scanner;
import java.util.InputMismatchException;

public class testClass {


    public static void main(String[] args) {

        String letter = new String();
        Scanner screen = new Scanner(System.in);
        boolean inputOK = false;
        String dump = null;
        while (!inputOK) {
            System.out.print("Enter ('y' or 'n': )");
            try {
                letter = screen.nextLine();
                dump = screen.nextLine();
                inputOK = true;
            } catch (InputMismatchException e) {
                dump = screen.nextLine();
                System.out.println("\"" + dump + "\" is not a legal letter, " +
                    "please try again!");
            } 
        } 
        screen.close();
        System.out.println("That is a valid letter");


    }
}

如果有人可以提供帮助，我们将不胜感激。 谢谢:)

Answer 1

首先，只会抛出InputMismatchException

to indicate that the token retrieved does not match the pattern for the expected type, or that the token is out of range for the expected type.

由于除了 y 和 n 之外的任何内容仍然是 String 的，因此不会抛出此错误。相反，如果不是 y 或 n，您可以抛出新的 InputMismatchException:

String letter = new String();
Scanner screen = new Scanner(System.in);
boolean inputOK = false;
while (!inputOK) {
      System.out.println("Enter ('y' or 'n': )");
      try {
          letter = screen.nextLine();

          if(!letter.equals("y") && !letter.equals("n")) {
              throw new InputMismatchException();
          }


          inputOK = true;
      } catch (InputMismatchException e) {

          System.out.println("\"" + letter + "\" is not a legal letter, " +
                "please try again!");

      } 
  } 
  System.out.println("That is a valid letter");

关闭System.in也不是一个好习惯。一般规则是如果您没有打开资源，则不应关闭它