Find the first occurrence with Regex and Java

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.

"some text" 38 m² "some text" ,

"some text" 48,8 m² "some text",

"some text" 48 m² "some text", etc..

What I have so far is:

\d\d,\d\s*(\m\u00B2)|\d\d\s*(\m\u00B2)

This right now finds all occurrences, although I guess it could be fixed with findFirst(). Any ideas how to improve the Regex part?

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.

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;
    }

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.

Find the first occurrence with Regex and Java

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