I created a test using grep but it does not work in sed.
grep -P '(?<=foo)bar' file.txt
This works correctly by returning bar.
sed 's/(?<=foo)bar/test/g' file.txt
I was expecting footest as output, but it did not work.
I created a test using grep but it does not work in sed.
grep -P '(?<=foo)bar' file.txt
This works correctly by returning bar.
sed 's/(?<=foo)bar/test/g' file.txt
I was expecting footest as output, but it did not work.
GNU sed does not have support for lookaround assertions. You could use a more powerful language such as Perl or possibly experiment with ssed which supports Perl-style regular expressions.
perl -pe 's/(?<=foo)bar/test/g' file.txt
Note that most of the time you can avoid a lookbehind (or a lookahead) using a capture group and a backreference in the replacement string:
sed 's/\(foo\)bar/\1test/g' file.txt
Simulating a negative lookbehind is more subtile and needs several substitutions to protect the substring you want to avoid. Example for (?<!foo)bar:
sed 's/#/##/g;s/foobar/foob#ar/g;s/bar/test/g;s/foob#ar/foobar/g;s/##/#/g' file.txt
# => ##).foobar here, => foob#ar or ba => b#a).foob#ar with foobar (or b#a with ba).## with #.Obviously, you can also describe all that isn't foo before bar in a capture group:
sed -E 's/(^.{0,2}|[^f]..|[^o].?)bar/\1test/g' file.txt
But it will quickly become tedious with more characters.
sed doesn't support lookarounds but choose (I'm the author) does. It uses PCRE2 syntax.
For example:
$ echo "hello bar foobar" | choose -r --sed '(?<=foo)bar' --replace test
hello bar footest
It's speed is comparable to sed.