I've been handed a script written by another and been asked to modify it according to their needs. There's one line that needs to be altered:
if [ "$user" == "nobody" ]
This needs to be expanded to include additional users that start with the letters p, u, and z. After the letter, all user accounts are a sequence of random numbers i.e. p33783201. I've been googling for hours now but my lack of understanding is starting to confuse me.
Thanks!
In a regex, "either the letter p, u, or z" is [puz], a number is [0-9], and a sequence of numbers is usually either [0-9]+ (one-or-more) if you're using a regex syntax that supports it or [0-9][0-9]* (zero-or-more) if you're forced to use POSIX Basix Regex which doesn't. Finally add ^...$ to require a full match (anchor the match to start & end of the string, instead of just anywhere within).
As you tagged the question with bash, assuming it's a #!/usr/bin/env bash script, you can use Bash's [[ with its regex match operator =~ (which uses POSIX Extended regex syntax):
if [[ $user == nobody || $user =~ ^[puz][0-9]+$ ]]; then
if [[ $user =~ ^(nobody|[puz][0-9]+)$ ]]; then
If it weren't Bash but a #!/bin/sh "POSIX shell" script, you could use expr for the same task (which is limited to the POSIX Basic dialect of regex and doesn't have +):
if [ "$user" = nobody ] || expr "$user" : "^[puz][0-9][0-9]*$" > /dev/null; then
or, even, grep (or one of the 20+ other regex-capable tools if performance is not a big concern):
if [ "$user" = nobody ] || (echo "$user" | grep -qs "^[puz][0-9][0-9]*$"); then
if echo "$user" | grep -Eqs "^(nobody|[puz][0-9][0-9]*)$"
Note that [ should use a single =, even though shells often accept ==.
