Using shell scripts, I use grep to find the word “error” in a log file:
grep error this.log.
How can I print or get the line 3 lines below the line that word “error” is located?
Thanks in advance for your response.

Use nawk or /usr/xpg4/bin/awk on Solaris.

With sed:

With shell:

hey please try the following one


grep -A <<n>> <<search pattern>> file_name
grep -B <<n>> <<search pattern>> file_name


NOTE: n is the number of lines you want to traverse before/after the search pattern .

radoulov,
You are very knowledgeable.
nawk 'x&&!--x;/error/{x=3}' file
is what I needed.
Thank you.

Hi. I am a new user to this site. This is actually my first post.

My question is similar. How can I grep for a particular string and have grep return that string along with the next 10 lines beneath the grepped string? manas_ranjan said to issue grep -A <<n>> <<search pattern>> file_name, but it doesn't look like -A is a grep option on the Solaris 10 box I am trying to run the command on. I assume I should install the latest version of GNU grep, but I am not an admin of this box, so that is not an option.

Are there alternatives such as awk/nawk or sed? If so, what would the exact commands be?

Thanks you very much in advance.

- LaLonde

Use nawk or /usr/xpg4/bin/awk on Solaris.

refer to radoulov's post he has given the example.

heres another alternative