In POSIX sh, string replacement is undefined. while trying to do PASS2="${PASS//[${special_chars}]/}"

Question

My Code is

special_chars='[=!=][=@=][=#=][=$=][=%=][=&=]'
PASS="e@0cc3auZeeSio&G"

PASS2="${PASS//[${special_chars}]/}"

I want PASS2 to have all the characters in PASS - special characters. This works fine, but there is shell check error on this.

PASS2="${PASS//[${special_chars}]/}"
             ^-- SC2039: In POSIX sh, string replacement is undefined.

I tried doing

PASS2=$(printf '%s' "$PASS2" | PASS//["${special_chars}"]/)

And

PASS2=$(printf '%s' "$PASS" | PASS//["${special_chars}"]/)

These does not work functionally.

Shellcheck is trying to tell you that the `//` construct in parameter expansion is not valid POSIX (it's a bashism). It is also not clear, what the special_chars shall do. Please tell us what PASS2 should be after the transformation. Do you want each of the characters !@$%& removed? — Jens, Apr 09 '20 at 14:51

score 0 · Accepted Answer · answered Apr 09 '20 at 15:00

0

This script passes shellcheck:

#!/bin/sh

special='!@#$%&'
PASS="e@0cc3auZeeSio&G"
PASS2=$(printf %s "$PASS" | tr -d "$special")
echo "$PASS2"

answered Apr 09 '20 at 15:00

Jens

Thanks a lot @Jens – Sashank Sekhar Singh Apr 09 '20 at 19:02
I used PASS2=$(echo "$PASS" | sed 's/[!@#$&%]//g') – Sashank Sekhar Singh Apr 09 '20 at 19:04
@SashankSekharSingh Wonderful. Yes, sed can also perform the character removal. The next step on Stackoverflow is to "accept" and/or upvote the answer that helped you most by clicking the check mark and/or the upward facing triangle above the vote count. – Jens Apr 09 '20 at 19:10

1 Answers1