how to display only the pattern

UNIX for Advanced & Expert Users
1

Hi,
I have a file with 500 Lines and I want to search for a pattern within this file which starts with sm_ and ends with ). However I just want to print the pattern only and not the entire line. How do I do it ?

Thanks,
p

2

Hi,

GNU grep has the -o option that does what you want, it will only print the pattern, not the entire line...

from the man page:
-o, --only-matching Show only the part of a matching line that matches PATTERN.

3

Not tested...

awk 'match($0,"sm_.*)"){print substr($0,RSTART,RLENGTH)}' $FILE
4

Using sed...

sed 's/^.*\(PATTERN\).*$/\1/' yourFile

5

does not work, since it will print all 500 lines of the file, not just those lines which contain the pattern. A possible solution using sed would be...

sed -n '/sm_.*)/{s/^.*\(sm_.*)\).*$/\1/;p;}' $FILE
6

I must be half asleep today :wink:

This is what you want...

sed -n 's/^.*\(PATTERN\).*$/\1/p' yourFile

or even better (in terms of performance)...

sed -n '/PATTERN/s/^.*\(PATTERN\).*$/\1/p' yourFile

So for the OPs specific example...

sed -n '/sm_.*)/s/^.*\(sm_.*)\).*$/\1/p' yourFile

(which is pretty much what you have Ygor)