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

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 2

Take a look at DecimalFormat that is a subclass of NumberFormat http://docs.oracle.com/javase/6/docs/api/java/text/DecimalFormat.html

DecimalFormat myFormatter = new DecimalFormat("###.###");

Answer 3

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 4

Use DecimalFormat with a format pattern string.

Answer 5

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 6

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)