Hello, I have a little trouble with grep...what i have is a file with many lines and any of the line may contain a word "error" , i want to grep this word "error" and print 5 lines above and 5 line below including the line in which "error" is present.... i can grep for the word error and print line which contains only error but how do i print 5 lines above and 5 line below of the line were "error" is found Appreciate your help Thx in advance Mehul

If it works for your version of grep the -A and -B flags specify lines before and after a pattern match. You can also do it with sed sed -n -e '/regexp/{=;x;1!p;g;$!N;p;D;}' -e h For other tricks with sed look here: http://www.student.northpark.edu/pemente/sed/sed1line.txt

Thx david it worked