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sed regex the last match

原文 2020-12-18 16:59:41 8 2 regex/ linux/ sed

I want to change in file in each line the last match to the XXXX Given the file format:

"192.xx2" "someting" "another2321SD"

I want to have:

"192.xx2" "someting" "XXXX"

only using sed My solution:

sed -r 's/"(?:.(?!".+"))+$/XXXX/' file

2 answers

Remember what was there before the last pair of double quotes in a capture.

sed -r 's/^(.*)"[^"]*"$/\1"XXXX"/'

Please bear in mind that

(?:...) is a non-capturing group and POSIX regex does not support this construct
(?....) is a negative lookahead and POSIX regex does not support this construct either.

To mask all content in between the last pair of double quotation marks , you can use

sed -r 's/(.*)".*"/\1"XXXX"/' file
sed 's/\(.*\)".*"/\1"XXXX"/' file

See the sed demo :

s='"192.xx2" "someting" "another2321SD"'
sed -r 's/(.*)".*"/\1"XXXX"/' <<< "$s"
# => "192.xx2" "someting" "XXXX"

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