Assume a text string my_string
$ my_string="foo bar=1ab baz=222;"
I would like to extract the alphanumeric string between keyword baz and the semi-colon.
How do I have to modify the following grep code using regex assertions to also exclude the trailing semi-colon?
$ echo $my_string | grep -oP '(?<='baz=').*'
222;
Unless the string that you want to extract may itself contain ;, the simplest thing is probably to replace . (which matches any single character) with [^;] (which matches any character excluding ;)
$ printf '%s\n' "$my_string" | grep -oP '(?<='baz=')[^;]*'
222
With grep linked to libpcre 7.2 or newer, you can also simplify the lookbehind using the \K form:
$ printf '%s\n' "$my_string" | grep -oP 'baz=\K[^;]*'
222
Those will print all occurrences in the string and assume the matching text doesn't contain newline characters (since grep processes each line of input separately).
Also easy with sed:
sed -n 's/.*baz=\([^;]*\).*/\1/p' <<< $my_string
222
Steeldriver's answer is accurate, but I have a hard time with lookaheads/behinds and would do it like this for readability (with bash):
my_string="foo bar=1ab baz=222;"
regex='baz=([0-9]+);'
[[ $my_string =~ $regex ]] &&
echo "${BASH_REMATCH[1]}"
With any POSIX shell:
For the text between the first occurrence of baz= and the last occurrence of ; after that:
my_string="foo bar=1ab baz=222;"
case $my_string in
(*baz=*\;*)
result=${my_string#*baz=}
result=${result%;*};;
(*) result=
esac
for the text between the first occurrence of baz= and the next occurrence of ; after that, replace % with %% above
for the last occurrence of baz=, replace # with ##.
foo baz=x; bar=y;? (xorx; bar=y?). And forbaz=x; baz=y;?