grep - show two parts of matching line

Question

I've read through the man page of grep and tried few things, none of them worked, at least not for me.

I want to extract a good readable line while tailing a log. This is a generic line in a log file I want to beautify:

26 Jan 2018 08:32:29,309 [TEXT] (myService-0) long.text.I.dont.care.about.but.is.different.in.every.line: [OTHERTEXT] Text im actually interested in

What I want is this:

26 Jan 2018 08:32:29,309 [TEXT] [OTHERTEXT] Text im actually interested in

I know that with grep -o -e ".*\[TEXT\]" I get the first part, and with grep -o -e "\[OTHERTEXT\].*", I get the second part.

But this is not displayed on one line, also not if I combine it into grep -o -e ".*\[TEXT\]" -e "\[OTHERTEXT\].*"

[TEXT] and [OTHERTEXT] always are there and are my 'separators', so can be used to support extracting the parts I need.

I initially thought I could use grep -o -e "(.*\[TEXT\]).*(\[OTHERTEXT\].*)" and then somehow use the matching groups $1 and $2, but either I don't see it or there is no way to do so.

Is there a way to achieve what I want?

Preferred is using grep (simply because I want to learn more about it), but if that is not possible then awk or sed are fine as well, it just has to be usable with a tail -f.

And I'm also open to other approaches to get to that point, so let me know what ways exist to get there.

Thanks, Tobias

Answer 1

You can use sed:

sed -E 's/(\[TEXT]).*(\[OTHERTEXT])/\1 \2/' file.log

26 Jan 2018 08:32:29,309 [TEXT] [OTHERTEXT] Text im actually interested in

This sed matches a pattern between [TEXT] and [OTHERTEXT] and captures them in 2 groups. In replacement it puts those markers back using back-refrences \1 \2

Answer 2

Using awk you could replace everything between ] and [ with ] [:

$ awk 'sub(/\].*\[/,"] [")' file
26 Jan 2018 08:32:29,309 [TEXT] [OTHERTEXT] Text im actually interested in