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

Question

0.00/5 (No votes)

See more:

C#

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"); }

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"); }

Posted 6-Apr-11 0:00am

Member 7751939

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Fredrik Bornander · Answer 1 · 2011-04-06T01:01:00

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

Java

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());
    }
  }
}

Hope this helps,
Fredrik Bornander