Grep the substring within the last occurrence of single quotes

Question

I am trying to grep a substring within a string inside the last occurance of single quotes.below is my String :

'abc''cde''efg'

Below is the command which I have used;

echo "'abc''cde''efg'" |rev|grep -m1 -oP "(?<=') .*?(?=')"

But I am not getting desired output. I am getting fge while it should efg .

Answer 1

If you reverse the whole string, you should also reverse the result after extraction.

However, using a PCRE regex (enabled via the P option) for you case can simplify the task:

grep -oP ".*'\K[^']+"

Here, .*' matches up to the last (rightmost) occurrence of ' , \\K match reset operator will discard the whole matched text and [^']+ will put into the result 1 or more chars other than ' .

Since not all systems offer the PCRE functionality with grep , you might consider an awk solution, too:

awk -F\'\' 'gsub(/^'"'"'+|'"'"'+$/, "", $0) {print $NF}'

Here, the string is split into fields with a '' substring ( -F\\'\\' ), then the leading/trailing ' are removed ( gsub(/^'"'"'+|'"'"'+$/, "", $0) ) and then the last field is printed ( print $NF ).

See the online demo .

Answer 2

Code Snippet :

sed -r "s/.*'(\w+)'$/\1/g"
efg

grep -oP ".*'\K[^']+(?=')"
efg