I need one regular expression for detect words with length 8 characters that contain 2+ numbers and 2+ characters (no special characters ).
I am near the solution and I did the regex on regex101.com .
The problem are the words that contains one number that not should be releveant for my regex.
I discarded all words with characters that contain min 7 characters with (?![A-Za-z]{7,}) .
I discarded all words with numbersthat contain min 7 numbers with (?![\d]{7,}) .
And I discarded the words that contain min 2 numbers and 2 characters (?=[a-zA-Z\d]{2})[A-Za-z\d]{8}.
Why vaff8loe is matched?
I created this regular expression because after I hae to replace the entire word with ******* . Like:
papave23 ciao il mio pin papaver1 è reeredji332ji con vaff8loe 1234567o 123t123t papavero 9o 123t123y
After with replace("regex","********")
********ciao il mio pin papaver1 è reeredji332ji con ******** 1234567o ******** papavero 9o ********
Use 2 zero-width positive lookaheads:
(?=.*?[a-zA-Z].*?[a-zA-Z]) Must contain 2 ASCII letters
(?=.*?[0-9].*?[0-9]) Must contain 2 digits
[a-zA-Z0-9]{8} Must be exactly 8 letters and/or digits
Add ^ and $ if not using matches() for running the regex.
That means a full regex of:
^(?=.*?[a-zA-Z].*?[a-zA-Z])(?=.*?[0-9].*?[0-9])[a-zA-Z0-9]{8}$
For best performance, replace the . pattern with a negative character class. In that case you might want to shorten it with a repeating non-capturing group:
(?=(?:[^a-zA-Z]*[a-zA-Z]){2})
(?=(?:[^0-9]*[0-9]){2})
UPDATE
As question was updated to say that regex is needed to replace such words with *'s, the ^ and $ anchors should be changed to \b word boundary patterns, and the negative character classes must be changed to only skip valid characters:
s = s.replaceAll("\\b(?=(?:[0-9]*[a-zA-Z]){2})(?=(?:[a-zA-Z]*[0-9]){2})[a-zA-Z0-9]{8}\\b", "********");
See regex101 for demo.
Note that vaff8loe in the given example only contains 1 digit, so should not be replaced.