Let's say I have two patterns 1111 and 6666 and one variable n (dynamic line numbers depending on file). Assume n=2 here, I want to print "2" lines after 1111 including the pattern 1111 and the second pattern line in the original order. The output should looks like,
1111
2222
3333--the above two line after first pattern
6666
6666
1111
2222
3333--the above two line after first pattern
6666
6666
I tried with grep -A2 -e '1111' -e '6666' TXT. It did the job partly right because it also prints two lines after the second pattern "6666", which are not wanted.
Thanks. I think perl is installed as default with Ubuntu 14. That's already very good to get job done with perl. Is that any possibility to use sed or awk? I suppose either sed or awk should be able to do this job. However, I cannot work it out.
Many thanks. You are always come with a super concise, super smart script to solve the problem. May I kindly ask a further question on this problem? How to modified this script if I want to print "2" lines before 1111 including the pattern 1111 and the second pattern line in the original order. Then the output should look like,
1111-->only one line because there no line above 6666
6666-->this line meets both patterns
9999-->two lines before the first pattern '1111'
1111
6666
6666
Indeed, I thought about this. What if I have to print the previous m lines before a third pattern? That means it is a mixed process of backward search with pattern 3 and two forward searches with pattern 1 and pattern 2. I searched google and found it's a hard job to print multiple previous lines with awk.