awk: output with lines before and after a match

Question

I've just started using bash and I've asked this question about an issue that I had using the loop 'while' with 'grep': grep issues when using two files - I've tried everything. I was told to use 'awk' instead of 'grep' in looping due to the following issues https://unix.stackexchange.com/questions/169716/why-is-using-a-shell-loop-to-process-text-considered-bad-practice Replacing 'grep' in loop for 'awk' indeed worked. However, I realized that I need an output that saves not only the matched line but also 1 line before and 2 lines after. I googled it, but I couldn't find a way to do it. Anyone knows how I can save lines after and before in 'awk'?

Thank you advance

Answer 1

Suppose you want the line before 29 and two after in the output of seq 40 (that outputs 40 lines counting from 1 when supported in your environment).
You can use awk with

seq 40 | awk 'printnext>0 {print; printnext--}
              /29/ {print previous;print; printnext=2;}
              {previous=$0}'

You should look what options your grep is supporting:

seq 40 | grep -B1 -A2 "29"