如何解析比NumberFormat在Java中更严格的数字？

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?

Eg I would like to not allow these three inputs, for an integer, but Numberformat allow all of them:

NumberFormat nf = NumberFormat.getNumberInstance();
nf.setParseIntegerOnly(true);

Number numberA = nf.parse("99.731");    // 99 (not what the user expect)
Number numberB = nf.parse("99s.231");   // 99 (invalid)
Number numberC = nf.parse("9g9");       // 9  (invalid)

System.out.println(numberA.toString());
System.out.println(numberB.toString());
System.out.println(numberC.toString());

Answer 1

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 2

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 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

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 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

将DecimalFormat与格式模式字符串一起使用。

Answer 7

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("###.###");