Shell Programming and Scripting

Normally grep command display the single line which matches the criterion. I want to display what one line before and after the matching lines in the list of files of a particular directory. Kindly help

man grep. look for -A or -B option.

ghostdog's answer assumes GNU grep, and just to further expand: if you want the first line before AND after, you can use -C. It's kind of convenient that you have -A, -B, and -C for these, but -A means "after", -B means "before" and -C can be thought of as "Context".

If you don't have access to GNU grep, you can use awk. Something like this awk script:

Code:

 /regexp/ { if (prev) print prev; print $0; prev=""; matched=1; next; }
 matched { print $0; matched = 0; next; }
 { prev=$0; }

i will give you some hints how to do it.

first check the files that contain the patten using -l option
grep -l patten *

Then in a loop use grep -n patten fileneme

to have the line numbers of the patten

example ==============

amit@softpc142108 ars $ grep -n passwdgen ars.txt
7:-rwxr-xr-x 1 amit smc 533 Apr 15 08:50 passwdgen
17:-rwxr-xr-x 1 amit smc 533 Apr 15 08:50 passwdgen


Here u can cut and pass the line number to a variable

grep -n passwdgen ars.txt | cut -d ':' -f1
7
17
=================================

then you can use sed to print the lines desired

sed -n '6,8p' filename.

awk

Code:

awk 'c&&c--;/CUSTOMER NAME/{c=1;print p}{p=$0}' file

Love amitranjansahu's solution...

Another thread awhile back similar in nature...

My solution -> https://www.unix.com/302298638-post10.html

(Original thread -> https://www.unix.com/unix-advanced-ex...-log-file.html )