Getting NumberFormatException

Question

if(Integer.parseInt(n)==(m[i].age))
            {
                m[i].showdetails();
            }

I am getting number format exception in if condition In this code n is an string and m[i].age is an integer and I want to compare them

thank you:)any help would be appreciated.

Well, now you know that `n` is not a valid number String. Your solution is to fix it so that it is. — Hovercraft Full Of Eels, Jun 17 '17 at 11:36
`n` is not a number. Did you try printing it before the statement to check its value? Hint: are you *100% sure* that `n` does not contain spaces? If it does, you need to remove them before parsing. — BackSlash, Jun 17 '17 at 11:36
@BackSlash: would the same exception be thrown if `m[i].age` is not a number? — Hovercraft Full Of Eels, Jun 17 '17 at 11:37
if you need more help from us, you should post the stacktrace as well as a valid [mcve] with your question. — Hovercraft Full Of Eels, Jun 17 '17 at 11:37

score -1 · Accepted Answer · answered Jun 17 '17 at 11:39

-1

dont try to convert string to integer instead convert integer to string

if(n.compareTo(Integer.toString(m[i].age))==0)
            {
                m[i].showdetails();
            }

answered Jun 17 '17 at 11:39

Saurabh Gupta

Thank you for the answer. Its working :) – Suraj Chauhan Jun 17 '17 at 11:51
1

That is one approach - but be aware that this way you'll just get `false` as a result for any `n` that is not exactly a valid representation of the given integer - you might miss bugs that way. Also, I'd prefer `n.equals(Integer.toString(m[i].age))` – Hulk Jun 17 '17 at 11:54
@SurajChauhan This is one approach, but it's wrong in my opinion. Always solve your bugs, **never** do hacks like this one to avoid solving your bugs. – BackSlash Jun 17 '17 at 12:25

1 Answers1