java - 我需要提供单独的异常语句。 1. 空字符串和 2. 有效数字数据

Question

try{
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
            System.out.println(" Enter the Amount of articles to be ordered.");
            amount = reader.readLine();

            if(amount.trim().isEmpty()){
                System.out.println("Amount Entered is Empty");
            }

            for(int count=0;count<amount.length();count++){
                if(!Character.isDigit(amount.charAt(count))){
                    throw new NumberFormatException();
                }
            }            
            order.validateAmount(amount);
        }catch(NumberFormatException numbere){
            System.out.println("Either Number format is uncorrect or String is Empty, Try Again.");
    }

上面的代码为我提供了针对空字符串异常和无效数字数据异常的单个 println() 语句，这是我不想要的。我想要两个异常的单独 println() 语句。如何获得？

Answer 1

1. 您可以使用两个不同的异常，例如 NumberFormatException 和 IllegalArgumentException 并执行两个不同的 catch-子句。

       ...
       if (amount.isEmpty())
           throw new IllegalArgumentException();
       ...
   
   } catch (NumberFormatException numbere) {
       System.out.println("Either Number format is uncorrect, Try Again.");
   } catch (IllegalArgumentException empty) {
       System.out.println("String is empty, Try Again.");
   }
   

2. 使用相同的异常，但使用不同的消息:

   try {
       BufferedReader reader = new BufferedReader(new InputStreamReader(
               System.in));
       System.out.println(" Enter the Amount of articles to be ordered.");
       String amount = reader.readLine();
   
       if (amount.trim().isEmpty()) {
           System.out.println("Amount Entered is Empty");
       }
   
       if (amount.isEmpty())
           throw new IllegalArgumentException("String is empty.");
   
   
       for (int count = 0; count < amount.length(); count++)
           if (!Character.isDigit(amount.charAt(count)))
               throw new IllegalArgumentException("Number format incorrect.");
   
       order.validateAmount(amount);
   } catch (IllegalArgumentException e) {
       System.out.println(e.getMessage() + " Try again.");
   }
   

3. 或者，您可以使用两个不同的构造函数推出自己的异常，并使用一个标志来说明异常是由于错误数字还是空字符串造成的.