always giving password not strong enough message (else condition is running) - bash

Question

I have been working on this function in shell scripting to take up a password from a user and check if it satisfies the criteria of being an effective password but it always says password not strong enough. The password I'm trying to use is LK#12abc.
Here is my code:
function paq()
{
read -p "PASSWORD :-" password
pasq="^(?=.*[A-Z].*[A-Z])(?=.*[!##$&*])(?=.*[0-9].*[0-9])(?=.*[a-z].*[a-z].*[a-z]).{8}$"
if [[ $password =~ $pasq ]]; then
echo "Valid password"
echo "The password is:- $password" >> user1.txt
echo "$password" >>password.txt
else
echo "password not strong enough"
fi
}
paq

Answer 1

It appears your password has to have:
3 lower case letters
2 upper case letters
2 digits
1 punctuation character
exactly 8 characters long
So, with bash glob patterns:
if [[ $password == *[a-z]*[a-z]*[a-z]* ]] &&
[[ $password == *[A-Z]*[A-Z]* ]] &&
[[ $password == *[0-9]*[0-9]* ]] &&
[[ $password == *[!##$\&*]* ]] &&
(( ${#password} == 8 ))
then
echo password OK
else
echo password does not satisfy criteria
fi
I'm surprised that I need to escape the & in the 4th test

Answer 2

As it says in man bash:
An additional binary operator, =~, is available, with the same
precedence as == and !=. When it is used, the string to the right of
the operator is considered an extended regular expression and matched
accordingly (as in regex(3)).
The first problem is that POSIX regex does not allow for lookahead
and lookbehing assertions.
The second problem is that your test password is not correct - you
miss 8 random characters at the end.
The third problem is that you didn't specify the
shebang and didn't quote variables correctly.
All in all, your script should look like that:
#!/usr/bin/env bash
paq() {
read -r -p "PASSWORD :- " password
pasq="^(.*[A-Z].*[A-Z])(.*[!##$&*])(.*[0-9].*[0-9])(.*[a-z].*[a-z].*[a-z]).{8}$"
if [[ "$password" =~ $pasq ]]; then
echo "Valid password"
echo "The password is:- $password" >> user1.txt
echo "$password" >>password.txt
else
echo "password not strong enough"
fi
}
paq
Run it:
$ ./paq.sh
PASSWORD :- AB!11abc12345678
Valid password

Answer 3

You can take your original regex and feed it to grep -P (only if your system's grep supports Perl regular expressions). Then use the script:
while ! echo "$password" | grep -P '^(?=.*[A-Z].*[A-Z])(?=.*[!##$&*])(?=.*[0-9].*[0-9])(?=.*[a-z].*[a-z].*[a-z]).{8}$'
do
read -r -p "PASSWORD: " password
echo -e "Password not strong enough, it needs\n\
3 lower case letters\n\
2 upper case letters\n\
2 digits\n\
1 punctuation character\n\
exactly 8 characters long\n"
done
echo "Valid password"
echo "The password is:- $password" | tee user1.txt

Answer 4

Using tr to translate each wanted character classes to a digit, sorting that digit string and count matching with grep (using #glennjackman defined character classes except the exact length):
echo LK#12abc |
tr '[:digit:]' 1 | # do this first
tr '[:upper:]' 2 |
tr '[:lower:]' 3 |
tr '[:punct:]' 4 |
tr -c '[:digit:]' 5 | # all else to 5
grep -o . | # transpose
sort | # sort
tr -d '\n' | # transpose
if
grep -q "1\{2,\}2\{2,\}3\{3,\}4\{1,\}5" # 5 is the trailing newline
then
echo yes
else
echo no
fi # outputs
yes
( I was trying to tr '[:digit:][:upper:][:lower:]' '123' but apparently that won't work so each class is now processed on its own line. Maybe you could loose the trs, just sort the password string and count match classes with grep)