Integer.Parse(String) throws exception

Question

I'm new at android developement, new to Java too. Originally, I'm a C# and C++ programmer, so this is kind of suspect to me. I wrote this code

String[] teile = temp.split(" ");

int[] teileInt = new int[] 
{
    50, 50, 50, 50, 50, 50
} ;

for (int i = 0; i < teile.length; ++i) {
    try {
        Integer integer = new Integer(teile[i]);
        //int d = Integer.parseInt(teile[i]);
    }
    catch (NumberFormatException ex)
    {
        Toast.makeText(this, teile[i] + getMessagesFromException(ex), Toast.LENGTH_SHORT).show();
    }
    //

}

Both expressions turn into the same Exception!

(Cant post an image)

The Toast says:

40null Invalid int: "40"
java.lang.NumberFormatException: Invalid
int: "40"

Im sorry, but I can' t figure out whats wrong... Is 40 not an Integer?

Your code does not show what the content of temp is, which constructs teile. — cYrixmorten
– cYrixmorten, Commented Mar 4, 2014 at 22:32
Post the relevant code where the exception arises. Also, post the content of temp String variable. — Luiggi Mendoza
– Luiggi Mendoza, Commented Mar 4, 2014 at 22:32

peter.petrov · Accepted Answer · 2014-03-04 22:36:02Z

2

Try this. Integer integer = new Integer(teile[i].trim());
I assume you have a leading or a trailing space in teile[i].

answered Mar 4, 2014 at 22:36

peter.petrov

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3 Comments

Andrei Nicusan Over a year ago

or better split by "\\s+"

peter.petrov Over a year ago

@AndreiNicusan Yes, that could help too. Agreed.

SonneXo Over a year ago

youre answer was really helpfull! didnt expect it would be neceressary to trim() those!!

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