Using the grep and cut delimiter command (in bash shell scripting UNIX) - and kind of "reversing" it?

So I have a file with the text:

puddle2_1557936:/home/

I want to use the grep command

grep puddle2_1557936

Mixed in with the cut command (or another command if neccessary) to display just this part:

/home/

So far, I know that if do this

 grep puddle2_1557936 | cut -d ":" -f1

then it will display

puddle2_1557936

So is there anyway to kind of "inverse" the delimiter cut command?

NOTE: The solution must start off with grep puddle2_15579636.

1 Answer

You don't need to change the delimiter to display the right part of the string with cut.

The -f switch of the cut command is the n-TH element separated by your delimiter : :, so you can just type :

 grep puddle2_1557936 | cut -d ":" -f2

Another solutions (adapt it a bit) if you want fun :

Using grep :

grep -oP 'puddle2_1557936:\K.*' <<< 'puddle2_1557936:/home/ /home/

grep -oP '(?<=puddle2_1557936:).*' <<< 'puddle2_1557936:/home/ /home/

or with perl :

perl -lne '/puddle2_1557936:(.*)/ and print $1' <<< 'puddle2_1557936:/home/ /home/

or using ruby (thanks to glenn jackman)

ruby -F: -ane '/puddle2_1557936/ and puts $F[1]' <<< 'puddle2_1557936:/home/ /home/

or with awk :

awk -F'puddle2_1557936:' '{print $2}' <<< 'puddle2_1557936:/home/ /home/

or with python :

python -c 'import sys; print(sys.argv[1].split("puddle2_1557936:")[1])' 'puddle2_1557936:/home/ /home/

or using only bash :

IFS=: read _ a <<< "puddle2_1557936:/home/" echo "$a" /home/

or using js in a shell :

js<<EOF var x = 'puddle2_1557936:/home/ print(x.substr(x.indexOf(":")+1)) EOF /home/

or using php in a shell :

php -r 'preg_match("/puddle2_1557936:(.*)/", $argv[1], $m); echo "$m[1]\n";' 'puddle2_1557936:/home/ /home/