Regular Expression Password Validation

Question

I'd like to use regular expression to validate the characters requirement of a password.

Requirement: Password should have 16 characters.

1. Character 1-4 should have at least 1 digit.
2. Character 5-8 should have at least 1 lower case character.
3. Character 9-12 should have at least 1 upper case character.
4. Character 13-16 should have at least 1 symbol (punctuation).

I've tried to use regular expression with a positive lookahead but it does not work finally:

echo 'XXXX9999ccccXXX%' | grep -P '^((?=.*[0-9]).{4})((?=.*[a-z]).{4})((?=.*[A-Z]).{4})((?=.*\pP).{4})$'

Answer 1

Your lookahead syntax is off, because it is not correctly checking the positions you mentioned in your requirements. The following regex pattern seems to work for me:

^(?=.{0,3}\d)(?=.{4,7}[a-z])(?=.{8,11}[A-Z])(?=.{12,15}[.,$%^&!@]).{16}$

Explanation:

(?=.{0,3}\d)           - number in positions 1-4
(?=.{4,7}[a-z])        - lowercase in positions 5-8
(?=.{8,11}[A-Z])       - uppercase in positions 9-12
(?=.{12,15}[.,$%^&!@]) - symbol in positions 13-16

Demo

I don't know grep or Linux well enough to comment on whether you are making best use, but this should at least fix any problems you were having with the pattern.