Number format exception

[vignesh Thamarai]

hello friends,
i have a hashset contain integers like 1,2,4,5 . now i want to get every integer value from hashset.

here is my code. set2 is a hashset.

   
 for(Iterator it1 = set2.iterator();it1.hasNext();)
        {
       Object element = it1.next();

       int i = Integer.parseInt(element.toString());              -------------> throws Number format exception here.

        grp = (User) Userlp.findById(i);

        as = new UserGroupmapping(usrId,grp,diObj);

        create(as);

        modified++;

        }

 

it throws a number format exception , i try this one this too not helping me

    int i = it1.next();   

please help me what iam missing...

[Campbell Ritchie]

What have you put into the Set in the first place? If there are Integers in it, there is no need to mess around with the parseInt and toString methods; you simply get the Integer out and can convert it to an int with auto-unboxing if necessary.

Your code is poorly formatted, which makes it difficult to read; there should, for example, be a single space after the semicolons in the for () line. Put the line starting int i = inside a try block, then follow it with a catch, so ti reads like this

 . . .
try
{
    int i = Integer.parseInt(element.toString());
}//end try
catch (NumberFormatException nfe)
{
    nfe.printStackTrace();
    System.err.println(element);
}//end catch

That will allow you to see what you have in your Set that isn't a number.

And why don't you parameterise the set as a Set<Integer>?