CODEWITHSUNDEEP
regexbashawkgrep
Joao Vitorino
Joao Vitorino

Reputation: 3256

Regex to extract part of line that match

I'm trying to write a regex to get the last word of each line of a file that start with "A" and end with "/"

grep '^A.*/$' list_svn.txt

This give the whole line:A Java/SYSTEM_A/Sources/branch/developerA/src/org/hl7/v3/but what I want is just "v3" string.

I tried awk but without success too.

grep '^A.*/$' list_svn.txt | awk -F '/' '{print $0, $(NF-1)}'
>> Java/SYSTEM_A/Sources/branch/developerA/src/org/hl7/v3/ v3

Upvotes: 1

Views: 2116

Answers (2)

Benjamin W.
Benjamin W.

Reputation: 52102

With GNU grep and look-arounds:

$ grep -Po '^A.*/\K[^/]*(?=/$)' <<< 'A  Java/SYSTEM_A/Sources/branch/developerA/src/org/hl7/v3/'
v3

-P enables Perl regular expressions (required for look-ahead and look-behind), and -o returns only the match.

The \K acts as a variable length look-behind: "match everything up to here but don't include it in the match". The actual match is a sequence of characters other than /, and the (?=/$) bit stands for "followed by a / and the end of the line" (but also not included in match).

Upvotes: 1

anubhava
anubhava

Reputation: 784898

You can avoid grep altogether as using awk you can do:

awk -F/ '/^A.*\/$/ {print $(NF-1)}' list_svn.txt

v3

Upvotes: 2

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