How to use grep to extract a substring?

Question

Possible duplicate:
extract regexp result from string and write it to variable

Here is my command:

grep -E '\ * [[: space:]] + FIN [[: space:]] + ([^)] +?)') myfile

It outputs:

• FIN (SUCCESS)

And I would like it to output only:

SUCCESS

How can I tell grep to do this?

Answer 1

You can pass the output of grep to awk.

 grep -E '*[[:space:]]+FIN[[:space:]]+([^)]+?)') myfile | awk '{print $2}' 

I'm not sure how to do this with grep alone, as it is not suitable for this particular use case. Since you are on a platform with grep, use the pipes to your advantage when you can have one team to solve part of the problem and another team in the other part.

Answer 2

grep cannot output one capture group, but you can use sed to do this:

 sed 's/\*[[:space:]]\+FIN[[:space:]]\+(\([^)]\+\))/\1/g' file 

Answer 3

If you use ack , you can use matching groups and the --output switch:

 ack '\*\s+FIN\((.+?)\)' --output='$1' myfile