sed command to display only the matched pattern and not the whole line

Hi,

I have a file with content as

2014 H_DSF_ABL.txt.gz
Dec 11 09:29 H_DSF_CDL.txt.gz
18 08:09 H_DSF_HCL.txt.gz
H_DSF_PPM.txt.gz

I need the output as

H_DSF_ABL.txt.gz
H_DSF_CDL.txt.gz
H_DSF_HCL.txt.gz
H_DSF_PPM.txt.gz

Please suggest the unix command to display the exact output as mentioned(only the pattern matched and not the whole line).
Also it should work on solaris OS.

Thanks!!

I suggest you to use sed. There are two different approach:
1. construct a regexp to find what you need, group it, and replace the whole line to that group
2. construct a regexp to find what you do not need and replace it with nothing.

what have you made so far?

You could match and delete everything up to the last space.
sed 's/.* //'

If this is all you need, “grep -o” might be a better option than using sed.

Originally Posted by mina86 If this is all you need, “grep -o” might be a better option than using sed.

Please elaborate. I tried this and couldn't make it work.

Edit: Ahh, now I understand. This works.

grep -o "H_DSF.*" $InFile >$OutFile

Daniel B. Martin

grep -Eo "[^[:space:]]+$"

?

Maybe this is a nitpick but I question this ...

sed 's/.* //'

... and this ...

grep -Eo "[^[:space:]]+$"

... because those solutions are based on position rather than pattern.

Daniel B. Martin

If you want the last element to be printed, even easier is

awk '{print $NF}'

Thank you all for your replies.

sed and awk command worked perfectly in unix and solaris OS.

sed 's/.* //'

awk '{print $NF}'

I got the desired output.
Actually grep -o is better option in unix and but it does not work in solaris.