How to parse numbers more strict than what NumberFormat does in Java?

Question

I\'m validating user input from a form.

I parse the input with NumberFormat, but it is evil and allow almost anything. Is there any way to parse number more strict?

Answer 1

Integer.parseInt(String) will throw a NumberFormatException on all of your examples. I'm not sure if that's what you're looking for, but it's definitely "more strict."

Answer 2

Maybe this helps:

String value = "number_to_be_parsed".trim();
NumberFormat formatter = NumberFormat.getNumberInstance();
ParsePosition pos = new ParsePosition(0);
Number parsed = formatter.parse(value, pos);
if (pos.getIndex() != value.length() || pos.getErrorIndex() != -1) {
    throw new RuntimeException("my error description");
}

(Thanks to Strict number parsing at mynetgear.net)

Answer 3

I wouldn't use java's number format routine, especially with the locale settings if you worry about validation.

    Locale numberLocale = new Locale(“es”,”ES");
    NumberFormat nf = NumberFormat.getInstance(numberLocale);
    ParsePosition pos = new ParsePosition(0);
    Number test = nf.parse("0.2", pos);

You would expect there to be an issue here, but no.. test is equal to 2 and pos has an index of 3 and error index of -1.

Answer 4

There are many ways to do that:

• regex - check if it matches("\\d+")
• with javax.validation - @Digits(fraction=0, integer=5)
• apache commons IntegerValidator

Answer 5

I gave up on writing my own validation class, and went with NEBULA WIDGETS FormattedText

It was written over the SWT widget API, but you can easily adapt the NumberFormatter class

Answer 6

Use DecimalFormat with a format pattern string.