Is it possible to make a replacement in an inverted way, where the regular expression is maintained and the other characters are eliminated?
Word
String word = "Lorem ipsum dolor sit amet, 27/08/2020 consectetur adipiscing elit. Etiam consequat nisi est.";
Regex
([0-9]{2}/[0-9]{2}/[0-9]{4})
word.replaceAll("([0-9]{2}/[0-9]{2}/[0-9]{4})","");
Expected
27/08/2020
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Answer
Yes, it’s called find().
String word = "Lorem ipsum dolor sit amet, 27/08/2020 consectetur adipiscing elit. Etiam consequat nisi est.";
Matcher m = Pattern.compile("([0-9]{2}/[0-9]{2}/[0-9]{4})").matcher(word);
if (m.find())
System.out.println(m.group());
Output
27/08/2020
Alternative, make the regex match everything, and use the capture group you already defined:
word.replaceFirst(".*?([0-9]{2}/[0-9]{2}/[0-9]{4}).*", "$1")
Result
27/08/2020
The main difference between the two if what happens of the regex doesn’t match, i.e. if the string doesn’t have a date in it.
The first will print nothing. You can explicitly control how you handle it by writing an else clause.
The second will result in the full original string, unchanged. You cannot control how you handle the “not found” condition.