Java Regex Number Validate parseNumber(String input)

Description

parse Number

License

Open Source License

Declaration

public static int parseNumber(String input) throws NumberFormatException 

Method Source Code

//package com.java2s;
//License from project: Open Source License 

import java.util.regex.Pattern;

public class Main {
    private static Pattern onlyNums = Pattern.compile("[^0-9]");
    private static Pattern containsNums = Pattern.compile(".*\\d.*");

    /**// w  w  w  .ja  v  a2  s. c  o  m
     * Returns a default value instead of a NumberFormatException when input is invalid
     *
     * @return matched number
     */
    public static int parseNumber(String input, int defaultReturn) {
        int num = defaultReturn;
        if (containsNumbers(input)) {
            input = stripNumber(input);
            num = Integer.parseInt(input);
        }
        return num;
    }

    public static int parseNumber(String input) throws NumberFormatException {
        int num;
        if (containsNumbers(input)) {
            input = stripNumber(input);
            num = Integer.parseInt(input);
        } else
            throw new NumberFormatException("Not a readable number \"" + input + "\"");
        return num;
    }

    public static boolean containsNumbers(String str) {
        return containsNums.matcher(str).matches();
    }

    public static String stripNumber(String input) {
        return onlyNums.matcher(input).replaceAll("");
    }
}

Related

1. isNumeric(String value)
2. isNumericAndCanNull(String src)
3. isNumericAndCanNull(String src)
4. isNumString(String txt)
5. parseNumber(String in)
6. parseNumber(String input)
7. parseNumber(String s)
8. parseNumbers(String str)