如何在JAVA中的差异原因对象上为相同的NumberFormatException打印差异消息? [英] How to print diff Msg for same NumberFormatException on diff cause objects in JAVA?

问题描述

How to print diff Msg for same NumberFormatException on diff cause objects in JAVA?

try {  int a=Integer.parseInt(aStr); int b= Integer.parseInt(bStr); }catch (NumberFormatException ex) {   if ex's cause is from int a;//ex.getCause()=a? System.out.println("a is not a integer");   if ex's cause is from int b System.out.println("b is not a integer"); }

尝试{int a = Integer.parseInt(aStr); int b = Integer.parseInt(bStr); } catch(NumberFormatException ex){如果ex的原因是来自int a;//ex.getCause()= a? System.out.println("a不是整数");}
catch(NumberFormatException ex){如果ex的原因来自int b System.out.println("b不是整数")； }

try { int a=Integer.parseInt(aStr); int b= Integer.parseInt(bStr); }catch (NumberFormatException ex) { if ex''s cause is from int a;//ex.getCause()=a? System.out.println("a is not a integer");}
catch (NumberFormatException ex){ if ex''s cause is from int b System.out.println("b is not a integer"); }

推荐答案

我会将整个内容包装在一个辅助方法中，如下所示:

I would wrap that whole thing up in a helper method, like this:

package com.test;

import java.text.ParseException;

public class Program {

  private static int CheckIntegerArgument(String value, String name) {
    try {
      return Integer.parseInt(value);
    }
    catch(NumberFormatException e) {
      throw new 
        IllegalArgumentException(String.format("%s is not a number (%s)", name, value));
    }
  }

  public static void someMethod(String a, String b) {
    int aValue =  CheckIntegerArgument(a, "a");
    int bValue =  CheckIntegerArgument(b, "b");

    // Use aValue and bValue here;
  }

  public static void main(String[] args) throws Exception, ParseException {
    try {
      someMethod("1234", "I am not a number");
    }
    catch(IllegalArgumentException e) {
      System.out.println(e.getMessage());
    }
  }
}

希望这会有所帮助，
弗雷德里克·博纳德(Fredrik Bornander)

Hope this helps,
Fredrik Bornander

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