String manipulation, optional character

Question

With grep, you can use a question mark ? to signify an optional character, that is a character that is to be matches 0 or 1 times.

$ foo=qwerasdf

$ grep -Eo fx? <<< $foo
f

The question is does Bash String Manipulation have a similar feature? Something like

$ echo ${foo%fx?}

Answer 1

You're probably talking about parameter expansion. It uses shell patterns, not regular expression, so the answer is no.

Upon further reading, I noticed that if you

shopt -s extglob

you can use extended pattern matching which can achieve something similar to regex, albeit with slightly different syntax.

Check this out:

word="mre"
# this returns true
if [[ $word == m?(o)re ]]; then echo true; else echo false; fi

word="more"
# this also returns true
if [[ $word == m?(o)re ]]; then echo true; else echo false; fi

word="mooooooooooore"
# again, true
if [[ $word == m+(o)re ]]; then echo true; else echo false; fi

Works with parameter expansion too,

word="noooooooooooo"
# outputs 'nay'
echo ${word/+(o)/ay}
# outputs 'nayooooooooooo'
echo ${word/o/ay}