Using grep and regex to get specific users list from etc/password

Question

I want to grab specific users ID from the /etc/passwd file in bash.

A given user ID is made of exactly 6 characters, either 6 integers or 1 letter and 5 integers (the letter is always the first character of the user ID in the second case). See example below:

398329
y32839
392009
r39288

I am looking for a regex pattern to get a list with only the user IDs that match this format, from my /etc/passwd file. This is what I did so far:

users="$(cut -d: -f1 /etc/passwd | grep -o '[0-9]*')"

I could get the user IDs with numbers but since my regex is limited, any help would be appreciated. Thanks!

Answer 1

You can use grep with correct regex like this:

cut -d: -f1 /etc/passwd | grep -E '^[a-zA-Z0-9][0-9]{5}$'
398329
y32839
392009
r39288

^[a-zA-Z0-9][0-9]{5}$ will match username of 6 digits or a letter followed by 5 digits.

As an improvement , you can replace cut + grep with this single awk command:

awk -F: '$1 ~ /^[a-zA-Z0-9][0-9]{5}$/{print $1}' /etc/passwd

Answer 2

grep with perl regex:

grep -oP '\w:\d{5}' /etc/passwd

example results:

4:65534
x:65534

I assume you're referring to the password encrypted field in etc/passwd, which is separated with colon; but if you really only concerned with UIDs that actually begin with a letter then simply remove the second colon in the regex, '\\w\\d{5}' .