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sed replace last line matching pattern

Tags:

regex

bash

sed

Given a file like this:

a
b
a
b

I'd like to be able to use sed to replace just the last line that contains an instance of "a" in the file. So if I wanted to replace it with "c", then the output should look like:

a
b
c
b

Note that I need this to work irrespective of how many matches it might encounter, or the details of exactly what the desired pattern or file contents might be. Thanks in advance.

like image 246
TTT Avatar asked Jun 14 '13 18:06

TTT


1 Answers

Not quite sed only:

tac file | sed '/a/ {s//c/; :loop; n; b loop}' | tac

testing

% printf "%s\n" a b a b a b | tac | sed '/a/ {s//c/; :loop; n; b loop}' | tac
a
b
a
b
c
b

Reverse the file, then for the first match, make the substitution and then unconditionally slurp up the rest of the file. Then re-reverse the file.

Note, an empty regex (here as s//c/) means re-use the previous regex (/a/)

I'm not a huge sed fan, beyond very simple programs. I would use awk:

tac file | awk '/a/ && !seen {sub(/a/, "c"); seen=1} 1' | tac
like image 87
glenn jackman Avatar answered Oct 07 '22 01:10

glenn jackman