CODEWITHSUNDEEP
regexperlsed
Ugur
Ugur

Reputation: 2044

Perl or sed: Regex replace with group capture

I cannot seem to make sed do a regex replace with group capture in this example:

What I tried with sed:

$ echo "DONE project 1" | sed -E 's/^DONE(.*)$/$1 DONE/g'
$1 DONE  # fail! no group capture in `sed`?

Is there a way to do this in Perl?

Actually this is part of an AppleScript used by DEVONthink[^1], and I realized that sed was not able to do regex search/replace with group capture.

[^1]: DEVONthink's search/replace script with AppleScript usage

set transformedName to do shell script "echo " & quoted form of itemName & " | sed -E 's/" & sourcePattern & "/" & destPattern & "/g'"
set name of selectedItem to transformedName

Upvotes: 2

Views: 610

Answers (2)

brian d foy
brian d foy

Reputation: 132822

The Perl version is very close to the sed version because Perl stole some of sed's features:

$ perl -pe 's/^DONE(.*)/\1 DONE/'

The -p effectively wraps while(<>) { ...; print } around your argument to -e.

Note that the /g flag doesn't make much sense here. You're going to match the first DONE and everything after it. There's no second match to make.

Upvotes: 2

Ryszard Czech
Ryszard Czech

Reputation: 18611

sed can easily do this, use \1, $1 is the Perl backreference syntax:

sed -E 's/^DONE(.*)/\1 DONE/g'

.* matches all the line up to the end, so you do not have to use $ in the LHS.

Upvotes: 2

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