如何在 Switch 语句中捕获用户输入的数字不存在的错误。

huangapple
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英文:

How to catch an Error in a Switch statement when the user entered number doesn't exist

问题

我试图使我的程序防错，该程序基本上是一个迷你计算器。但我不知道如何编写一个“Catch”语句，以便在用户输入不存在的情况下（在我的情况下为任何负数或大于4的数字）进行检测。

  1. public static void main(String[] args) {
  2.     System.out.println("你好用户！你想要使用哪种操作？");
  3.     System.out.println("1) + \n2) - \n3) * \n4) /");
  4.     
  5.     Scanner operacijai = new Scanner(System.in);
  6.     int operacija = operacijai.nextInt();
  7.     
  8.     int n = 1;
  9.     do {
  10.         try {
  11.             switch (operacija) {
  12.                 case 1:
  13.                     addingMethod();
  14.                     n = 2;
  15.                     break;
  16.                     
  17.                 case 2:
  18.                     subtractingMethod();
  19.                     n = 2;
  20.                     break;
  21.                     
  22.                 case 3:
  23.                     multiplyingMethod();
  24.                     n = 2;
  25.                     break;
  26.                     
  27.                 case 4:
  28.                     dividingMethod();
  29.                     n = 2;
  30.                     break;
  31.             }
  32.         } catch (Exception e) {
  33.             System.out.print("请输入正确的数字！");
  34.         }
  35.     } while (n == 1);
  36.     operacijai.close();
  37. }
英文:

I'm trying to error proof my program that basically works as a mini calculator. But I have no idea how to write a "Catch" statement that would detect when the user enters a case number that doesn't exist, in my case anything that is negative or > 4

  1. 		System.out.println("Hello user! Which operation would you like to use?");
  2. 		System.out.println("1) + \n2) - \n3) * \n4) /");
  3. 			
  4. 		
  5. 		Scanner operacijai = new Scanner(System.in);
  6. 		int operacija = operacijai.nextInt();
  7. 		
  8. 		
  9. 		int n=1;
  10. 		do {
  11. 		try {
  12. 			switch (operacija) {
  13. 			case 1:
  14. 				addingMethod();
  15. 				n=2;
  16. 				break;
  17. 					
  18. 			case 2:
  19. 				subtractingMethod();
  20. 				n=2;
  21. 				break;
  22. 				
  23. 			case 3:
  24. 				multiplyingMethod();
  25. 				n=2;
  26. 				break;
  27. 				
  28. 			case 4:
  29. 				dividingMethod();
  30. 				n=2;
  31. 				break;
  32. 					} 		
  33. 			}
  34. 			catch(Exception e) {
  35. 				System.out.print("Enter a correct number!");
  36. 			}
  37. 	
  38. 		} while(n==1);
  39. 		operacijai.close();
  40. 		
  41. 	} ```
  43. </details>
  46. # 答案1
  47. **得分**: 2
  49. 为什么你要不必要地抛出一个 `Exception`？我建议你在 `switch` 中加入一个带有所需错误消息的 `default` 情况。另外，将输入部分移动到循环内，这样它就会继续获取输入。
  51. 我还建议你使用 `nextLine()` 而不是 `nextInt()`。了解更多信息，请查看 https://stackoverflow.com/questions/13102045/scanner-is-skipping-nextline-after-using-next-or-nextfoo。
  53. ```java
  54. import java.util.Scanner;
  56. public class Main {
  57.     public static void main(String[] args) {
  58.         System.out.println("Hello user! Which operation would you like to use?");
  59.         System.out.println("1) + \n2) - \n3) * \n4) /");
  60.         Scanner operacijai = new Scanner(System.in);
  61.         int operacija = 0, n = 1;
  62.         boolean valid;
  63.         do {
  64.             do {
  65.                 valid = true;
  66.                 try {
  67.                     operacija = Integer.parseInt(operacijai.nextLine());
  68.                 } catch (NumberFormatException e) {
  69.                     System.out.println("Enter an integer only.");
  70.                     valid = false;
  71.                 }
  72.             } while (!valid);
  73.             switch (operacija) {
  74.                 case 1:
  75.                     System.out.println("addingMethod()");
  76.                     n = 2;
  77.                     break;
  79.                 case 2:
  80.                     System.out.println("subtractingMethod()");
  81.                     n = 2;
  82.                     break;
  84.                 case 3:
  85.                     System.out.println("multiplyingMethod()");
  86.                     n = 2;
  87.                     break;
  89.                 case 4:
  90.                     System.out.println("dividingMethod()");
  91.                     n = 2;
  92.                     break;
  93.                 default:
  94.                     System.out.println("Invalid input");
  95.             }
  96.         } while (n == 1);
  97.     }
  98. }

一个示例运行:

  1. Hello user! Which operation would you like to use?
  2. 1) + 
  3. 2) - 
  4. 3) * 
  5. 4) /
  6. 5
  7. Invalid input

另一个示例运行:

  1. Hello user! Which operation would you like to use?
  2. 1) + 
  3. 2) - 
  4. 3) * 
  5. 4) /
  6. a
  7. Enter an integer only.
  8. 5
  9. Invalid input
  10. 2
  11. subtractingMethod()
英文:

Why do you want to throw an Exception unnecessarily? I suggest you just put a default case in your switch with the required error message. Also, move the input part inside the loop, so that it continues to take input.

I also suggest you use nextLine() instead of nextInt(). Check https://stackoverflow.com/questions/13102045/scanner-is-skipping-nextline-after-using-next-or-nextfoo to learn more about it.

  1. import java.util.Scanner;
  3. public class Main {
  4. 	public static void main(String[] args) {
  5. 		System.out.println(&quot;Hello user! Which operation would you like to use?&quot;);
  6. 		System.out.println(&quot;1) + \n2) - \n3) * \n4) /&quot;);
  7. 		Scanner operacijai = new Scanner(System.in);
  8. 		int operacija = 0, n = 1;
  9. 		boolean valid;
  10. 		do {
  11. 			do {
  12. 				valid = true;
  13. 				try {
  14. 					operacija = Integer.parseInt(operacijai.nextLine());
  15. 				} catch (NumberFormatException e) {
  16. 					System.out.println(&quot;Enter an integer only.&quot;);
  17. 					valid = false;
  18. 				}
  19. 			} while (!valid);
  20. 			switch (operacija) {
  21. 			case 1:
  22. 				System.out.println(&quot;addingMethod()&quot;);
  23. 				n = 2;
  24. 				break;
  26. 			case 2:
  27. 				System.out.println(&quot;subtractingMethod()&quot;);
  28. 				n = 2;
  29. 				break;
  31. 			case 3:
  32. 				System.out.println(&quot;multiplyingMethod()&quot;);
  33. 				n = 2;
  34. 				break;
  36. 			case 4:
  37. 				System.out.println(&quot;dividingMethod()&quot;);
  38. 				n = 2;
  39. 				break;
  40. 			default:
  41. 				System.out.println(&quot;Invalid input&quot;);
  42. 			}
  43. 		} while (n == 1);
  44. 	}
  45. }

A sample run:

  1. Hello user! Which operation would you like to use?
  2. 1) + 
  3. 2) - 
  4. 3) * 
  5. 4) /
  6. 5
  7. Invalid input

Another sample run:

  1. Hello user! Which operation would you like to use?
  2. 1) + 
  3. 2) - 
  4. 3) * 
  5. 4) /
  6. a
  7. Enter an integer only.
  8. 5
  9. Invalid input
  10. 2
  11. subtractingMethod()

答案2

得分: 0

你还可以处理在default中的使用情况。完全取决于您的用例，您可以如下处理异常，还可以创建自定义异常并从default中抛出，类似于：

  1. System.out.println("Hello user! Which operation would you like to use?");
  2. System.out.println("1) 加法\n2) 减法\n3) 乘法\n4) 除法");
  4. Scanner operacijai = new Scanner(System.in);
  5. int operacija = operacijai.nextInt();
  7. int n=1;
  8. do {
  9.     try {
  10.         switch (operacija) {
  11.             case 1:
  12.                 addingMethod();
  13.                 n=2;
  14.                 break;
  16.             case 2:
  17.                 subtractingMethod();
  18.                 n=2;
  19.                 break;
  21.             case 3:
  22.                 multiplyingMethod();
  23.                 n=2;
  24.                 break;
  26.             case 4:
  27.                 dividingMethod();
  28.                 n=2;
  29.                 break;
  30.             default:
  31.                 System.out.print("输入正确的数字！");
  32.                 throw new CustomException();
  33.         }       
  34.     }
  35.     catch(CustomException e) {
  36.         System.out.print("输入正确的数字！");
  37.     }
  39. } while(n==1);
  40. operacijai.close();

注意：这里的代码只是翻译了您提供的部分内容，并没有对代码逻辑进行任何更改。

英文:

You can also handle the use case in default
It is totally upto your use-case how you are handling the exception, you can also create your custom exception and throw from default
something like:

  1.     System.out.println(&quot;Hello user! Which operation would you like to use?&quot;);
  2.     System.out.println(&quot;1) + \n2) - \n3) * \n4) /&quot;);
  5.     Scanner operacijai = new Scanner(System.in);
  6.     int operacija = operacijai.nextInt();
  9.     int n=1;
  10.     do {
  11.     try {
  12.         switch (operacija) {
  13.         case 1:
  14.             addingMethod();
  15.             n=2;
  16.             break;
  18.         case 2:
  19.             subtractingMethod();
  20.             n=2;
  21.             break;
  23.         case 3:
  24.             multiplyingMethod();
  25.             n=2;
  26.             break;
  28.         case 4:
  29.             dividingMethod();
  30.             n=2;
  31.             break;
  32.         default:
  33.             System.out.print(&quot;Enter a correct number!&quot;)
  34.             throw new CustomException();
  35.         }       
  36.       }
  37.       catch(CustomException e) {
  38.           System.out.print(&quot;Enter a correct number!&quot;);
  39.       }
  41.     } while(n==1);
  42.     operacijai.close();
  44. }

答案3

得分: 0

已找到一种使用默认情况的清晰方法。

  1. public static void main(String[] args) {
  2.     System.out.println("你好！请选择要使用的操作：");
  3.     System.out.println("1) 加法\n2) 减法\n3) 乘法\n4) 除法");
  5.     Scanner operacijai = new Scanner(System.in);
  6.     int operacija;
  8.     do {
  9.         operacija = operacijai.nextInt();
  10.         switch (operacija) {
  11.             case 1:
  12.                 addingMethod();
  13.                 break;
  15.             case 2:
  16.                 subtractingMethod();
  17.                 break;
  19.             case 3:
  20.                 multiplyingMethod();
  21.                 break;
  23.             case 4:
  24.                 dividingMethod();
  25.                 break;
  27.             default:
  28.                 System.out.print("请输入正确的数字！");
  29.         }
  31.     } while (operacija < 1 || operacija > 4);
  32.     operacijai.close();
  33. }
英文:

Figured out a clean way of doing this with default case.

  1. 		System.out.println(&quot;Hello user! Which operation would you like to use?&quot;);
  2. 		System.out.println(&quot;1) + \n2) - \n3) * \n4) /&quot;);
  3. 			
  4. 		
  5. 		Scanner operacijai = new Scanner(System.in);
  6. 		int operacija;
  7. 		
  8. 		do {
  9. 			operacija = operacijai.nextInt();
  10. 			switch (operacija) {
  11. 			case 1:
  12. 				addingMethod();
  13. 				break;
  14. 					
  15. 			case 2:
  16. 				subtractingMethod();
  17. 				break;
  18. 				
  19. 			case 3:
  20. 				multiplyingMethod();
  21. 				break;
  22. 				
  23. 			case 4:
  24. 				dividingMethod();
  25. 				break;
  26. 				
  27. 			default:
  28. 					System.out.print(&quot;Enter a correct number!&quot;);
  29. 				
  30. 					} 		
  31. 	
  32. 		} while(operacija &lt; 1 || operacija &gt; 4);
  33. 		operacijai.close();
  34. 		
  35. 	}
  37. </details>

huangapple
  • 本文由 发表于 2020年4月6日 02:54:49
  • 转载请务必保留本文链接：https://java.coder-hub.com/61047815.html
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