Remove all text before last space in text file from CLI

Question

I want to remove everything before the last space in a text file from the CLI

Example text:

2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar

The string that should remain is dump-2015-03-02-21-34.tar

Solution 1

There's a lot of choice using the command line:

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | cut -c32-
dump-2015-03-02-21-34.tar

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | sed 's/.* //'
dump-2015-03-02-21-34.tar

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | grep -oP "\S+$"
dump-2015-03-02-21-34.tar

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | awk {'print $NF'}
dump-2015-03-02-21-34.tar

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | perl -pe 's/.* //'
dump-2015-03-02-21-34.tar

$ echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | rev | cut -d ' ' -f 1 | rev
dump-2015-03-02-21-34.tar

Solution 2

sed is by default greedy, so just remove everything until the last space:

sed 's/.* //'

or use this awk idiom:

awk '$0=$NF'

Solution 3

Try grep:

echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | grep -Eo "[^ ]+$"

You can also use tr and cut together:

echo "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar" | tr -s ' ' | cut -d' ' -f4

Why should python be left behind:

python2 -c 'print "2015-03-02 21:34:15      20480 dump-2015-03-02-21-34.tar".split()[3]'

All of the above produce output: dump-2015-03-02-21-34.tar