pattern search and display the last occuring pattern?

Question

I have a log file containing startup and shutdown times for everday.

I want to see the last pattern pattern for startup and shutdown(which is located at the end of the file being updated everyday).

Also I dont want all the lines between the startup and shutdown.I want only few(say 3) lines after the last startup and few(say 3 lines) after the last shutdown.

Any suggestions for a one liner.

Answer 1

You can use a combination of sed and grep if you don't mind to reverse the (extracted) lines of the log file twice (see How can I reverse the lines in a file?).

# test case
echo '
startup 1
shutdown 1
startup 2
shutdown 2
startup 3
line 1
line 2
line 3
line 4
line 5
shutdown 3
line 1
line 2
line 3
line 4
line 5
' | 
tail -r | sed -n -e '1,/startup/p' | tail -r | grep -E -A 3 '(startup|shutdown)'


# output
startup 3
line 1
line 2
line 3
--
shutdown 3
line 1
line 2
line 3

Answer 2

try to use grep instead sed wit option '--after-context'.

Example:

cat logfile | grep --after-context=3 startup