Invalid int when converting from string to integer

Question

I'm trying to convert a string that contains just numbers to integer using int number = Integer.parseInt(string), but it returns #

error: Invalid int: "number"

For example, if the string is "10", it returns: Invalid int: "10". What is wrong?

Edit:

FileInputStream fis;
        int number =0;
        String line="";
        try
        {
            fis = new FileInputStream(getFilesDir()+pathToFile);
            BufferedReader reader = new BufferedReader(new InputStreamReader(fis));
            line=reader.readLine();
            Log.e("Read Line", "Read line: *" +line+ "*"); //returns *10*
            try
            {
                number = Integer.parseInt(line.trim());
            }
            catch(NumberFormatException nfe)
            {
                Log.e("ERROR CONVERTING", nfe.getMessage()); // return above error
            }
            reader.close();

        }
        catch(Exception ex)
        {
            Log.e("ERROR OPENING FILE", "Can't open file: "+ex.getMessage());
        }

might be white space around string. try Integer.parseInt(string.trim()); — David M
– David M, Commented Feb 6, 2013 at 14:30
Are you sure it's "ten" but not "one o"? I mean, check if zero is a zero, not caps o — Yaroslav Mytkalyk
– Yaroslav Mytkalyk, Commented Feb 6, 2013 at 14:30
I suspect the string isn't just "10". Try logging its length as well - there may be some non-printable characters. — Jon Skeet
– Jon Skeet, Commented Feb 6, 2013 at 14:31

Community · Accepted Answer · 2017-05-23 12:14:49Z

4

I used the code in the answer from this question: LINK

str = str.replaceAll("\\D+","");

Don't know why it doesn't work without this, because there are no other characters besides numbers in the string, but not it works.

edited May 23, 2017 at 12:14

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answered Feb 6, 2013 at 18:16

domen

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Comments

Charleston · Accepted Answer · 2017-09-26 16:10:09Z

2

Some times it means "the number is too big".

I come across a situation where it happens because my number was too big. Example:

Integer.parseInt("201506121234567");

It causes me the exception NumberFormat.

To solve, I changed to this:

Long.parseLong("201506121234567");

Hope it helps someone!

answered Sep 26, 2017 at 16:10

Charleston

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Comments

Fritz · Accepted Answer · 2013-02-06 14:31:36Z

1

Did you actually try with "10" or any other number? The error is pretty specific.

Invalid int: "number"

This means that in this line

Integer.parseInt(string)

string has the value "number". Check where are you setting this value and verify that you're actually setting a numeric string.

answered Feb 6, 2013 at 14:31

Fritz

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3 Comments

m0skit0 Over a year ago

From OP: if the string is "10", it returns: Invalid int: "10"

Fritz Over a year ago

@m0skit0 Actually, the OP started that sentence with For example. I'm not too sure he actually tried that.

Fritz Over a year ago

Then I'd use a regex to actually depure the string and find the numeric characters on it.

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