CODEWITHSUNDEEP
javaregex
TechGeek
TechGeek

Reputation: 508

Regex to extract four digits using Java Pattern

I'm trying to extract four digits before the file extension using Java Pattern Matchers. It's throwing no group found exception. Can someone help me on this ?

String fileName = "20210101-000000_first_second_1234.csv";
Pattern pattern = Pattern.compile("\\\\d{4}");
System.out.println(pattern.matcher(fileName).group(4));

I would like to get 1234 from the fileName. I compiled the file pattern using regex \\\\d{4}. Which returns four groups. So, fourth group should suppose to return 1234 which is not returning, instead throwing group not found exception.

Upvotes: 2

Views: 117

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626748

The "\\\\d{4}" string literal defines a \\d{4} regex that matches a \dddd string (a backslash and then four d chars). You try to access Group 4, but there is no capturing group defined in your regex. Besides, you can't access match groups before actually running the matcher with Matcher#find or Matcher#matches.

You can use

String fileName = "20210101-000000_first_second_1234.csv";
Pattern pattern = Pattern.compile("\\d{4}(?=\\.[^.]+$)");
Matcher m = pattern.matcher(fileName);
if (m.find()) {
    System.out.println(m.group());
}

See the Java demo and the regex demo. Details:

  • \d{4} - four digits
  • (?=\.[^.]+$) - a positive lookahead that requires a . char and then one or more chars other than . till end of string.

Note also the Matcher m = pattern.matcher(fileName) added and if (m.find()) checks if there is a match. Only if there is a match, the value can be retrieved from the 0th group, m.group().

Upvotes: 6

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