Using regex with IFS variable to split Bash string

Question

I am trying to split a string into an array by any character that is not alphanumeric. Can assign a regex pattern to the IFS variable to accomplish this?

I have tried it like so:

input="$1"

IFS="[^a-zA-Z]"
read -ra name_parts <<< "$input"

But this splits the string by any "a" or "A" - not even recognizing the "^". This question looks similar by title, but does not appear to be about the question I'm asking.

Answer 1

IFS cannot be used that way. It does not take a regular expression. At the minimum, the characters (literal) composing the IFS is used by the shell to split words when it does expansion of words. E.g.

IFS=: read -r v1 v2 <<<"foo:bar"

What you have defined in IFS="[^a-zA-Z]" takes the characters literally i.e. each of [, ^, a, -, z, A, Z and ] are used as separators to split your input string which is clearly not something you would expect to do.

Answer 2

IFS is just a bunch of characters (or bytes), not a regex. But you could use e.g. awk or sed to split the string based on a regex, print it out with a simpler separator and then read it with the shell's read.

read -ra name_parts < <(awk -vFS='[^a-zA-Z]' -vOFS=' ' '{$1=$1; print}' <<< "$input")

or

read -ra name_parts < <(sed -e 's/[^a-zA-Z]/ /g' <<< "$input")

Answer 3

Instead of tinkering with IFS, you're better off mapping the the input string and then splitting it using the default IFS:

read -ra name_parts <<<"$(printf '%s\n' "$input" | LC_ALL=C tr -cs 'a-zA-Z\n'  '[ *]')"

Now the array name _parts will hold the string sliced at the non alphabetic positions.