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
– anubhava, Commented Nov 4, 2014 at 15:04
@anubhava He's looking for a fully anchored replacement. That finds a match anywhere. — Etan Reisner
– Etan Reisner, Commented Nov 4, 2014 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
– Etan Reisner, Commented Nov 4, 2014 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
– James Brown, Commented Nov 4, 2014 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
– Chris Dodd, Commented Nov 4, 2014 at 15:33

Community · Accepted Answer · 2017-05-23 11:43:54Z

5

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"
$

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answered Nov 4, 2014 at 15:06

fedorqui

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