I have a text file where every line is in a similar format to this:
%#&#%# [Dinero] / Money /
!#@%$@ [Dia] / Day /
$%&$^#@ [Perro] / Dog /
I am looking to extract the words inside the brackets, ie. Ola, Dinero, Perro, etc, and save it all to a new text file line by line. Essentially, I am looking to omit/delete/erase all words, letters, special characters, and anything else outside the brackets, including the brackets themselves.
with the help of awk,
$ awk -F'[][]' '{print $2}' < input
Dinero
Dia
Perro
Using grep,
grep -oP '\[\K[^\]]+' input
\K means that use look around regex advanced feature. More precisely, it's a positive look-behind assertion
if you lack the -P option, you can do this with perl:
perl -lne '/\[\K[^\]]+/ and print $&' input
use -i option to edit file in place.
Or simply you can use cut as suggested by @juliepelletier,
cut -d"[" -f2 < input | cut -d"]" -f1
cat input|cut -d\[ -f2|cut -d\] -f1
Commented
Jul 4, 2016 at 18:13
grep -o -e '\[.*\]' file.txt | sed -e 's/\[//' -e 's/\]//' to find and extract all words with the brackets surrounding them and then strip away the brackets. And yes, that's the slightly verbose variant - could also be done more concisely with the same effect. It just better describes the flow I'd have put the data through. I keep meaning to pick up awk but I can never find the time.
sed 's/^.*\[//;s/\].*$//' /path/to/input > /path/to/output
