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;
/*//from  w ww.j ava 2 s  .c o m
 * This file is part of
 * KeepXP Server Plugin for Minecraft
 *
 * Copyright (C) 2013 Diemex
 *
 * KeepXP is free software: you can redistribute it and/or modify
 * it under the terms of the GNU Affero Public License as published by
 * the Free Software Foundation, either version 3 of the License, or
 * (at your option) any later version.
 *
 * KeepXP is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU Affero Public License
 * along with KeepXP.  If not, see <http://www.gnu.org/licenses/>.
 */

import java.util.regex.Pattern;

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

    public static int parseNumber(String input) throws NumberFormatException {
        int num;
        input = onlyNums.matcher(input).replaceAll("");
        if (input.length() > 0)
            num = Integer.parseInt(input);
        else
            throw new NumberFormatException("Not a readable number \"" + input + "\"");
        return num;
    }
}

Related

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