如何以更具體的方式處理NumberFormatException？

Question

我想以更具體的方式處理NumberFormatException。發生 此異常，當它試圖分配什麼，但一個整數，輸入以下時：

1. 串
2. 字符
3. 空輸入
4. 雙號

根據什麼被輸入我想顯示一個適當的消息，如

你輸入的字符串，請輸入整數

或

值不能爲空，請輸入一個整數值

下面的代碼捕獲NumberFormatException一般辦法。

我不知道是否有辦法包含更多的catch子句。

import java.util.Scanner; 

    public class TestException { 

     static int input; 
     static Scanner scan = new Scanner(System.in); 

     public static void main(String[] args) { 
      System.out.println("Enter an integer number: "); 

      try { 
       input = Integer.parseInt(scan.next()); 
       System.out.println("You've entered number: " + input); 
      } catch (NumberFormatException e) { 
       System.out.println("You've entered non-integer number"); 
       System.out.println("This caused " + e); 
      } 
     } 
    } 

Answer 1

您對使用if-else結構來catch塊內指定scenerios。

請參見下面的代碼：

String inString = null; 
try 
{ 
    iString = scan.next().trim(); 
    input = Integer.parseInt(inString); 
    System.out.println("You've entered number: " + input); 
} 
catch (NumberFormatException e) 
{ 
    if(inString.equals("") 
    { 
     System.out.println("You've entered empty string."); 
    } 
    else if(inString.length() == 1) 
    { 
     System.out.println("You've entered a single char"); 
    } 
    else 
    { 
     System.out.println("You've entered non-intereger number"); 
    } 
    System.out.println("This caused " + e); 
} 

Answer 2

首先從用戶獲取用戶輸入後，嘗試將其轉換爲整數。

static int input; 
    static Scanner scan = new Scanner(System.in); 

    public static void main(String[] args) { 
     System.out.println("Enter an integer number: "); 

     String inputString = scan.next(); 

     try { 
      input = Integer.parseInt(); 
      System.out.println("You've entered number: " + input); 
     } catch (NumberFormatException e) { 
      if(inputString.equals("") || inputString == null) { 
       System.out.println("empty input"); 
      } else if(inputString.length == 1) { 
       System.out.println("char input"); 
      } else { 
       System.out.println("string input"); 
      } 
     } 
    } 

Answer 3

你可以做輸入一些測試，如果解析輸入一個整數值出現異常，這樣的事情：

String scanned = null 
try { 
    scanned = scan.next(); 
    input = Integer.parseInt(scanned); 
    System.out.println("You've entered number: " + input); 
} catch (NumberFormatException e) { 
    if (scanned == null || scanned.isEmpty()) { 
     System.out.println("You didn't enter any value"); 
    } else if (scanned.length() == 1) 
     System.out.println("You entered a single char which is not a number"); 
    } 
    // and more tests, you can even try to parse as Double 
} 

Answer 4

String aString = null; 
    aString = scan.next().trim(); 
    System.out.println("You've entered number: " + aString); 
    if("".equals(aString.trim())){ 
     System.out.println("You have entered an Empty String"); 
    }else if(!isNumber(aString) && aString.length()==1){ 
     System.out.println("You have entered a Character"); 
    }else if(!isNumber(aString) && aString.length()>1){ 
     System.out.println("You have entered a String"); 
    }else if(isNumber(aString)){ 
     int input = Integer.parseInt(aString.replaceAll(",","")); 
     System.out.println("You have entered a correct Number"+input); 
    } 

    private boolean isNumber(String s){ 
     return s.matches("[0-9]+(,[0-9]+)*,?"); 
    }