Question

I have the following regex that checks for a password policy. it is validated to work:

(^([zZ]\d{3})*$)|((?=.{9,})(?=.*?[^\w\s])(?=.*?[0-9])(?=.*?[A-Z]).*?[a-z].*)

I want to use the regex in a bash script to validate a psssword the following way :

echo $password | grep "(^([zZ]\d{3})*$)|((?=.{9,})(?=.*?[^\w\s])(?=.*?[0-9])(?=.*?[A-Z]).*?[a-z].*)"
if [[ $? -eq 0 ]] ; then

This does not work in bash.
My question is :

How do i convert this “pure” regex to a one that work in bash ? which characters do i need to escape, what is the correct way to pass the regex to grep ? are there any other things i need to look out for ?

Thanks

Answer 1

This may be difficult.

Standard grep has limited capabilities. It only supports POSIX extended regular expressions which do not recognize the lookahead assertions your regex relies on.

If you have GNU grep on your machine, you can pass it the -P or --perl-regexp parameter, allowing it to use Perl compatible regular expressions. Then your regex should work.

As mentioned in my comment, the regex as-is is not suitable for passwort validation. It allows passwords like z000 or even the empty string:

(                 # Either match and capture...
 ^                #  Start of the string
 (                #  Match (and capture, uselessly in this case)
  [zZ]            #  case-insensitive z
  \d{3}           #  three digits
 )*               #  zero(!) or more times
 $                #  until the end of the string
)                 # End of first group.
|                 # OR
(                 # match and capture...
  (?=.{9,})       #  a string that's at least 9 characters long,
  (?=.*?[^\w\s])  #  contains at least one non-alnum, non-space character,
  (?=.*?[0-9])    #  at least one ASCII digit
  (?=.*?[A-Z])    #  at least one ASCII uppercase letter
  .*?[a-z].*      #  and at least one ASCII lowercase letter
)                 # no anchor to start/end of string...

Better use

^(?=.{9})(?=.*?[^\w\s])(?=.*?[0-9])(?=.*?[A-Z]).*?[a-z].*$