Find the first occurrence with Regex and Java

Question

I would like to be able to find the first occurrence of m² and then numbers in front of it, could be integers or decimal numbers. E.g.

"som

Answer 1

To get the first match, you just need to use Matcher#find() inside an if block:

String rx = "\\d+(?:,\\d+)?\\s*m\\u00B2";
Pattern p = Pattern.compile(rx);
Matcher matcher = p.matcher("E.g. : 4668,68 m² some text, some text 48 m²  etc");
if (matcher.find()){
    System.out.println(matcher.group());
}

See IDEONE demo

Note that you can get rid of the alternation group using an optional non-capturing group (?:..)?

Pattern breakdown:

• \d+ - 1+ digits
• (?:,\d+)? - 0+ sequences of a comma followed with 1+ digits
• \s* - 0+ whitespace symbols
• m\u00B2 - m2.

Answer 2

This is what I came up with you help :) (work in progress, later it should return BigDecimal value), for now it seems to work:

 public static String findArea(String description) {

        String tempString = "";
        Pattern p = Pattern.compile("\\d+(?:,\\d+)?\\s*m\\u00B2");

        Matcher m = p.matcher(description);

        if(m.find()) {
            tempString = m.group();
        }
//remove the m and /u00B2 to parse it to BigDecimal later
        tempString = tempString.replaceAll("[^0-9|,]","");
        System.out.println(tempString);
        return tempString;
    }

Answer 3

One simple way of doing it!

description.replaceFirst(@NotNull String regex,
                           @NotNull String replacement)

JAVADoc: Replaces the first substring of this string that matches the given regular expression with the given replacement.