Regex password validation in Bash shell

Question

I am using Regexes in Bash Shell script. I am using the below Regex code to check password criteria : Password should be at least 6 characters long with at least one digit and at least one Upper case Alphabet. I validated in the Regex validation tools, the Regex I have formed works fine. But, it fails in Bash Shell Script. Please provide your thoughts.

echo "Please enter password for User to be created in OIM: "
echo "******Please Note: Password should be at least 6 characters long with one digit and one Upper case Alphabet******"
read user_passwd
regex="^(?=.*[A-Z])(?=.*[a-z])(?=.*\d)\S{6,}$"
echo $user_passwd
echo $regex
if [[ $user_passwd =~ $regex ]]; then
    echolog "Password Matches the criteria"
else
    echo "Password criteria: Password should be at least 6 characters long with one digit and one Upper case Alphabet"
    echo "Password does not Match the criteria, exiting..."
    exit
fi

A single regular expression isn't necessarily the best solution for this kind of problem. As you can see from anubhava's answer, a series of tests that correspond directly to your requirements can be simpler and easier to maintain. — Keith Thompson
– Keith Thompson, Commented Dec 30, 2013 at 19:01
Your regexp checks for at least one lower case letter, but you don't mention that requirement in your question. — Keith Thompson
– Keith Thompson, Commented Dec 30, 2013 at 19:02

anubhava · Accepted Answer · 2013-12-30 19:44:27Z

5

BASH regex engine doesn't support lookarounds as in your regex.

You can use following shell glob check to make sure password matches your criteria:

[[ ${#s} -ge 6 && "$s" == *[A-Z]* && "$s" == *[a-z]* && "$s" == *[0-9]* ]]

It will make sure that input string $s meets all these conditions:

at least 6 characters long
has at least one digit
has at least one Upper case Alphabet
has at least one Lower case Alphabet

edited Dec 30, 2013 at 19:44

answered Dec 30, 2013 at 18:51

anubhava

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9 Comments

Keith Thompson Over a year ago

Just for the sake of clarity, I'd write >= 6 rather than > 5. BTW, the question didn't mention a requirement for at least one lower case letter.

anubhava Over a year ago

Yes >= 6 is better. I saw OP's regex "^(?=.*[A-Z])(?=.*[a-z])(?=.*\d)\S{6,}$" that was checking for lowercase character as well.

anubhava Over a year ago

In my testing both 123AsaA and 123AsaA came as valid passwords.

Shankar Guru Over a year ago

Hi Anubhava, Thank you so much for the answer. New Learning for me here. I have one doubt here (hope I dont sound too dumb) I entered range of inputs i.e. the length of the input (ranging from 6-12), the above code works between the length of input between 6-10 but after 10, the control moves to else condition and fails. Please let me know if there is any specific reason for it? Input provided: Success : aas12A AsaA123 123AssaA AAAsss123 Failed: aaaSSSAAA1 aaaSSSAAA112

anubhava Over a year ago

ok can you post the code you tried in your question then I can certainly test it out and suggest you accordingly.

|

dozer · Accepted Answer · 2018-06-28 12:01:52Z

1

I'm adding to anubhava's answer. His check worked for me but not completely. For e.g. "hello123" or "HELLO123" also passed the check, so it didn't detect case. The issue was with the locale settings, more specifically the LC_COLLATE variable. It needs to be set to "C" to consider case. But a better solution was to use character classes rather than range expressions. The explanation here helped me to solve my problem. Eventually using character classes worked for me.

[[ ${#s} -ge 6 && "$s" == *[[:lower:]]* && "$s" == *[[:upper:]]* && "$s" == *[0-9]* ]]

answered Jun 28, 2018 at 12:01

dozer

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