grep last match and it's following lines

Question

I've learnt how to grep lines before and after the match and to grep the last match but I haven't discovered how to grep the last match and the lines underneath it.

The scenario is a server log. I want to list a dynamic output from a command. The command is likely to be used several times in one server log. So I imagine the match would be the command and somehow grep can use -A with some other flag or variation of a tail command, to complete the outcome I'm seeking.

Answer 1

using awk instead:

awk '/pattern/{m=$0;l=NR}l+1==NR{n=$0}END{print m;print n}' foo.log

small test, find the last line matching /8/ and the next line of it:

kent$  seq 20|awk '/8/{m=$0;l=NR}l+1==NR{n=$0}END{print m;print n}'
18
19

Answer 2

The approach I would take it to reverse the problem as it's easier to find the first match and print the context lines. Take the file:

$ cat file
foo
1
2
foo
3
4
foo
5
6

Say we want the last match of foo and the following to lines we could just reverse the file with tac, find the first match and n lines above using -Bn and stop using -m1. Then simple re-reverse the output with tac:

$ tac file | grep foo -B2 -m1 | tac
foo
5
6

Tools like tac and rev can make problems that seem difficult much easier.