Saturday, April 7, 2012

java regex -- replace with captured substring but modified

Any time you have a string with lots of x.xx000001 or x.xx99999, it's probably noise you want to get rid of. Here's a java solution. Perl can do this in 1 line (at most 2).

    public static String cleanUp999or000(String orig) {
        final static Pattern PATTERN9999 = Pattern
                .compile("(\\d\\.\\d*)([0-8]9999+)(\\d\\s)");
        Matcher m = PATTERN9999.matcher(orig);
        StringBuffer sb = new StringBuffer();
        String without999 = orig;
        String without999_or_000 = orig;
        try {
            while (m.find()) {
                final long intEndingIn999 = Long.parseLong(m.group(2));
                final long intEndingIn000 = intEndingIn999 + 1;
                System.out.println(intEndingIn000);
                m.appendReplacement(sb, m.group(1) + intEndingIn000 + m.group(3));
            }
            m.appendTail(sb);
            without999 = sb.toString();
        } catch (NumberFormatException e) {
            e.printStackTrace();
            without999 = orig;
        } finally {
            without999_or_000 = without999.replaceAll(
                "(\\d\\.\\d+?)0000+\\d(\\s)", "$1$2");
        }
        return without999_or_000;
    }

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