Quick sed lesson

I was looking for a way to "grep" logs within certain time range. The first one I found (and use all the time since then) is using awk:-

awk '/run_cmd/ && "00:00:00" < $3 && $3 <= "04:35:00"' /var/log/local1

But what if I want just logs from 5 minutes ago? From certain time to now is easy:-

awk '/ERROR/ && "10:40:00" < $3 && $3 <= strftime("%H:%M:%s")' /var/log/local1

But I can't find a way to get "time ago" with awk. Of course we can use date command to generate the date and pass it to awk:-

time_ago=$(date --date='10 minutes ago' '+%H:%M') && awk -v time_ago="$time_ago" '/ERROR/ && time_ago < $3 && $3 <= strftime("%H:%M:%s")' /var/log/local1

That starting to get out of hand however. As usual, stackoverflow has better answer. Not using awk but sed.

sed -n "/^$(date --date='5 minutes ago' '+%b %_d %H:%M')/,\$p" /var/log/local1

While it work, I don't really understand it. Time for some quick sed lesson.

Basic sed - sed -n 10,15p file.txt print line 10 - 15 of the file.

10,15 is an address range. Address range can be a line number or regex pattern.

sed -n /^813/p file.txt this function similar to grep ^813 file.txt.

Oh, p there is a command, which is print. Another command usually use is d for delete.

Remember that address can be a range?

sed -n /^813/, \$p file.txt

Here we specify a range, which simply the end of the file. So sed will print starting from the match to the end of the file.

sed -n "/^$(date --date='5 minutes ago' '+%b %_d %H:%M')/,\$p" /var/log/local1

Now back to the original command. date --date='5 minutes ago' '+%b %_d %H:%M' will return something like Oct 12 10:14.

So basically we're doing:-

sed -n "/^Oct 12 10:14/,\$p" /var/log/local1

Which mean print from match of Oct 12 10:14 to the end of file.

Oh wait, why the -n flag there? sed by default will print each line. So -n is for quiet so we instruct it to not print anything. Instead we use p to print only the match.

End of lesson ;)