Disapprove password if it contains exactly 4 digits (validation)


I can’t figure out how to modify my regex to make sure the password follows the last condition:

  • at least 2 capital letters in a row
  • doesn’t have space symbols
  • contains digits
  • doesn’t contain 4 consecutive digits

{4} It currently disapproves the password if it has 4 and more digits but I need it to disapprove the password with EXACTLY 4 digits.

s1 = 'annII#443'
s2 = 'annII#4343'
s3 = 'annII#43434'

pattern = r"^(?=.*[A-Z]{2,})(?=.*[A-Za-z])(?=.*[0-9])(?!.*[0-9]{4})(?!.*[s]).*$"

re.findall(pattern, s1) # ['aШnnII#443']
re.findall(pattern, s2) # []
re.findall(pattern, s3) # []

PS: It’s just a task so don’t worry. It’s not gonna be used for any real purposes.

Asked By: sash a



  1. To match exactly four digits, you can use at start ^ e.g. (?!(?:.*D)?d{4}(?!d)).
    This requires start or a D non-digit before the 4 digits and disallows a digit after.
  2. (?=.*[A-Za-z]) looks redundant if you already require (?=.*[A-Z]{2,}) (2 upper).
  3. {2,} two or more is redundant. {2} would suffice and does not change the logic.
  4. Instead of (?!.*[s]).*$ you can just use S*$ (upper matches non-whitespaces).
  5. It’s generally more efficient to use lazy .*? or even a negated class where possible.

See this demo at regex101 (added n to negations in multiline demo for staying in line)

Answered By: bobble bubble
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