Java library to check whether a String contains a

问题:

I'm looking for a method that returns a boolean if the String it is passed is a valid number (e.g. "123.55e-9", "-333,556"). I don't want to just do:

public boolean isANumber(String s) {
    try { 
        BigDecimal a = new BigDecimal(s); 
        return true;
    } catch (NumberFormatException e) {
        return false;
    }
}

Clearly, the function should use a state machine (DFA) to parse the string to make sure invalid examples don't fool it (e.g. "-21,22.22.2", "33-2"). Do you know if any such library exists? I don't really want to write it myself as it's such an obvious problem that I'm sure I'd be re-inventing the wheel.

Thanks,

Nick

回答1:

I would avoid re-inventing this method and go with Apache Commons. If your using Spring, Struts or many other commonly used java libraries, they often have Apache commons included. You will want the commons-lang.jar file. Here is the method in NumberUtils you would want:

isNumber[1]

public static boolean isNumber(java.lang.String str)
Checks whether the String a valid Java number.

Valid numbers include hexadecimal marked with the 0x qualifier, scientific notation and numbers marked with a type qualifier (e.g. 123L).

Null and empty String will return false.

Parameters:
str - the String to check
Returns:
true if the string is a correctly formatted number

回答2:

use a regexp

回答3:

The exact regular expression is specified in the Javadocs for Double.valueOf(String).

To avoid calling this method on an invalid string and having a NumberFormatException be thrown, the regular expression below can be used to screen the input string:

final String Digits     = "(\\p{Digit}+)";
final String HexDigits  = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally 
// signed decimal integer.
final String Exp        = "[eE][+-]?"+Digits;
final String fpRegex    =
       ("[\\x00-\\x20]*"+  // Optional leading "whitespace"
        "[+-]?(" + // Optional sign character
        "NaN|" +           // "NaN" string
        "Infinity|" +      // "Infinity" string

        // A decimal floating-point string representing a finite positive
        // number without a leading sign has at most five basic pieces:
        // Digits . Digits ExponentPart FloatTypeSuffix
        // 
        // Since this method allows integer-only strings as input
        // in addition to strings of floating-point literals, the
        // two sub-patterns below are simplifications of the grammar
        // productions from the Java Language Specification, 2nd 
        // edition, section 3.10.2.

        // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
        "((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+

        // . Digits ExponentPart_opt FloatTypeSuffix_opt
        "(\\.("+Digits+")("+Exp+")?)|"+

        // Hexadecimal strings
        "((" +
        // 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
        "(0[xX]" + HexDigits + "(\\.)?)|" +

        // 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
        "(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +

        ")[pP][+-]?" + Digits + "))" +
        "[fFdD]?))" +
        "[\\x00-\\x20]*"); // Optional trailing "whitespace"

if (Pattern.matches(fpRegex, myString))
    Double.valueOf(myString); // Will not throw NumberFormatException
else {
    // Perform suitable alternative action
}

回答4:

Here is a regexp based utility function working fine (couldn't fit the "" check in the regexp while keeping it readable):

public class TestRegexp {
    static final String NUM_REGEX=
        "-?((([0-9]{1,3})(,[0-9]{3})*)|[0-9]*)(\\.[0-9]+)?([Ee][0-9]*)?";
    public static boolean isNum(String s) {
            return s!=null && s.length()>0 && s.matches(NUM_REGEX);  
    }
    public static void main(String[]args) {
        String[] values={
                "",
                "0",
                "0.1",
                ".1",
                "-.5E5",
                "-12,524.5E5",
                "-452,456,456,466.5E5",
                "-452,456,456,466E5",
                "22,22,2.14123415e1",
        };
        for (String value : values) {
            System.out.println(value+" is a number: "
            +isNum(value));
        }
    }

}

回答5:

Yeah a regular expression should do the trick. I only know .Net regexp but all regex languages are fairly similar so this should get you started. I didn't test it so you might want to kick it around a bit with the Java regex class.

"-?(([0-9]{1,3}(,[0-9{3,3})*)|[0-9]*)(\.[0-9]+(e-?[0-9]*)?)?"

Some of the Regex control syntax:
? - Optional element
| - OR operator. Basically I allowed numbers with or without commas if they were formatted correctly.
[ ] - Set of allowed characters
{ , } - Minimum maximum of element
* - Any number of elements, 0 to infinity
+ - At least one element, 1 to infinity
\ - Escape character
. - Any character (Hence why it was escaped)