Bash and replacing the whole string in variable?

Question

I'm trying to use Bash replacement to replace the whole string in variable if it matches a pattern. For example:

pattern="ABCD"

${var/pattern/}

removes (replaces with nothing) the first occurrence of $pattern in $var

${var#pattern}

removes $pattern in the beginning of $var

But how can I remove regex pattern "^ABCD$" from $var? I could:

if [ $var == "ABCD" ] ; then 
  echo "This is a match." # or whatever replacement
fi

but that's quite what I'm looking for.

What's wrong with `"${var/$pattern/}"`? Can you clarify with some example inputs. — anubhava, Nov 04 '14 at 15:04
@anubhava He's looking for a fully anchored replacement. That finds a match anywhere. — Etan Reisner, Nov 04 '14 at 15:05
I can't think of anything materially better than that `if` statement. `[ "$var" = ABCD ] && var=""` is slightly more terse but the same idea. — Etan Reisner, Nov 04 '14 at 15:06
@etan and @ fedorgui it works for me. Out of curiosity I'm going to leave this open for a while to see if there are other solutions. Thanks guys. — James Brown, Nov 04 '14 at 15:14
Note that your first two examples do NOT modify `$var` -- they expand to a string that comes from `$var` and has the corresponding change, but `$var` itself remains unchanged -- later instances of `$var` will not be affected. — Chris Dodd, Nov 04 '14 at 15:33

score 5 · Accepted Answer · edited May 23 '17 at 11:43

You can do a regular expression check:

pattern="^ABCD$"
[[ "$var" =~ $pattern ]] && var=""

it checks $var with the regular expression defined in $pattern. In case it matches, it performs the command var="".

$ pattern="^ABCD$"
$ var="ABCD"
$ [[ "$var" =~ $pattern ]] && echo "yes"
yes
$ var="1ABCD"
$ [[ "$var" =~ $pattern ]] && echo "yes"
$

1 Answers1