如何在Java中使用Scanner读取字符串？

huangapple
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英文:

How do I get the Scanner in java to read a string?

问题

  1. import java.util.*;
  3. public class Main {
  5.     public static void main(String[] args) {
  7.         String input;
  8.         Scanner scan = new Scanner(System.in);
  9.         
  10.         System.out.println("Enter your age, or enter 'q' to quit the program.");
  11.         input = scan.next();
  12.         
  13.         if (input.equalsIgnoreCase("q")) {
  14.             return;
  15.         }
  17.         int age = Integer.parseInt(input);
  18.         
  19.         System.out.println("Your age is " + age);
  20.     }
  21. }
英文:

How would I get my program to quit when the user enters q?
Is there something wrong with the scanner?

My code

  1. import java.util.*;
  2. public class Main{
  4.          public static void main(String []args){
  5.              
  6.              int age;
  7.              
  8.             Scanner scan = new Scanner(System.in);
  9.             System.out.println("Enter your age, or enter 'q' to quit the program.");
  10.             age = scan.nextInt();
  11.             
  12.             
  13.             if(age.equals("q") || age.equals("Q")){
  14.                 
  15.                 return 0;
  16.                 
  17.             }
  18.             
  19.             
  20.             
  21.             System.out.println("Your age is " + age);
  22.             
  23.             
  24.             
  25.          }
  26.     
  27.     }

答案1

得分: 0

我在你的代码中主要看到两个问题：

  1. 代码缺少一个循环来重复询问年龄。有许多种方法（for、while、do-while）可以编写循环，但我认为对于这种情况，do-while 是最合适的，因为它始终至少执行do块中的语句一次。
  2. age 的类型是 int，因此无法与字符串进行比较，例如你的代码 age.equals("q") 是不正确的。处理这种情况的一种好方法是将输入存储在类型为 String 的变量中，并在需要时检查其值，以确定是否应该允许/禁止处理它（例如尝试将其解析为 int）。

注意，当你尝试解析一个无法解析为 int 的字符串（例如 "a"）时，会得到一个 NumberFormatException，你需要处理它（例如显示错误消息，更改某些状态等）。

  1. import java.util.Scanner;
  3. public class Main {
  4.     public static void main(String[] args) {
  5.         int age;
  6.         String input;
  7.         Scanner scan = new Scanner(System.in);
  8.         boolean valid;
  10.         do {
  11.             // 假设输入是有效的
  12.             valid = true;
  13.             System.out.print("输入你的年龄，或输入 'q' 退出程序: ");
  14.             input = scan.nextLine();
  16.             if (!(input.equals("q") || input.equals("Q"))) {
  17.                 try {
  18.                     // 尝试将输入解析为整数
  19.                     age = Integer.parseInt(input);
  20.                     System.out.println("你的年龄是 " + age);
  21.                 } catch (NumberFormatException e) {
  22.                     System.out.println("无效输入");
  23.                     // 将 valid 的值更改为 false
  24.                     valid = false;
  25.                 }
  26.             }
  27.         } while (!valid || !(input.equals("q") || input.equals("Q")));
  28.     }
  29. }

一个示例运行：

  1. 输入你的年龄，或输入 'q' 退出程序: a
  2. 无效输入
  3. 输入你的年龄，或输入 'q' 退出程序: 12.5
  4. 无效输入
  5. 输入你的年龄，或输入 'q' 退出程序: 14
  6. 你的年龄是 14
  7. 输入你的年龄，或输入 'q' 退出程序: 56
  8. 你的年龄是 56
  9. 输入你的年龄，或输入 'q' 退出程序: q
英文:

I can see mainly two problems in your code:

  1. It lacks a loop to repeat asking for age again. There can be many ways (for, while, do-while) to write a loop but I find do-while most appropriate for such a case as it always executes the statements within the do block at least once.
  2. age is of type, int and therefore it can not be compared with a string e.g. your code, age.equals("q") is not correct. A good way to handle such a situation is to get the input into a variable of type, String and check the value if it should allow/disallow processing it (e.g. trying to parse it into an int).

Note that when you try to parse a string which can not be parsed into an int (e.g. "a"), you get a NumberFormatException which you need to handle (e.g. show an error message, change some state etc.).

  1. import java.util.Scanner;
  3. public class Main {
  4. 	public static void main(String[] args) {
  5. 		int age;
  6. 		String input;
  7. 		Scanner scan = new Scanner(System.in);
  8. 		boolean valid;
  10. 		do {
  11. 			// Start with the assumption that input will be valid
  12. 			valid = true;
  13. 			System.out.print("Enter your age, or enter 'q' to quit the program: ");
  14. 			input = scan.nextLine();
  16. 			if (!(input.equals("q") || input.equals("Q"))) {
  17. 				try {
  18. 					// Try to parse input into an int
  19. 					age = Integer.parseInt(input);
  20. 					System.out.println("Your age is " + age);
  21. 				} catch (NumberFormatException e) {
  22. 					System.out.println("Invalid input");
  23. 					// Change the value of valid to false
  24. 					valid = false;
  25. 				}
  26. 			}
  27. 		} while (!valid || !(input.equals("q") || input.equals("Q")));
  28. 	}
  29. }

A sample run:

  1. Enter your age, or enter 'q' to quit the program: a
  2. Invalid input
  3. Enter your age, or enter 'q' to quit the program: 12.5
  4. Invalid input
  5. Enter your age, or enter 'q' to quit the program: 14
  6. Your age is 14
  7. Enter your age, or enter 'q' to quit the program: 56
  8. Your age is 56
  9. Enter your age, or enter 'q' to quit the program: q

huangapple
  • 本文由 发表于 2020年10月17日 05:33:40
  • 转载请务必保留本文链接：https://go.coder-hub.com/64396673.html
  • integer
  • java
  • java.util.scanner
  • loops
  • string
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