print last field from line + alternative for awk

Question

Due to technical reason on my Solaris machine, I can't use awk in order to print the last field in line.

What are the other alternatives to awk that print the last field from line (using cut or tr ...etc)?

Example 1:

/usr/bin/hostname
machine1b
/usr/bin/hostname  | /usr/bin/sed 's/\(.\{1\}\)/\1 /g' |  /usr/bin/awk  '{print $NF}'
b

Example2

echo 1 2 3 4 5 | /usr/bin/awk  '{print $NF}'

5

Answer 1

Perl

echo foo bar baz | perl -pe 's/.*[ \t]//'

If you have to strip trailing spaces first, do it like this:

echo "foo bar baz " | perl -lpe 's/\s*$//;s/.*\s//'

The following was contributed by mr.spuratic in a comment:

echo "foo bar baz " | perl -lane 'print $F[-1]'

Bash

echo foo bar baz | while read i; do echo ${i##* }; done

or is bash is not your default shell:

echo foo bar baz | bash -c 'while read i; do echo ${i##* }; done'

If you have to strip a single trailing space first, do

echo "foo bar baz " | while read i; do i="${i% }"; echo ${i##* }; done

tr and tail

echo foo bar baz | tr ' ' '\n' | tail -n1

although this will only work for a single line of input, in contrast to the solutions above. Suppressing trailing spaces in this approach:

echo "foo bar baz " | tr ' ' '\n' | grep . | tail -n1

Answer 2

Using Cut:

Last field:

echo 1 2 3 4 5 | rev | cut -f1 -d ' '

Last character:

echo 1 2 3 4 5 | rev | cut -c1

Answer 3

echo hello my friend | tac -s' ' | tr '\n' ' ' | cut -d' ' -f 1

Answer 4

Something like this should work:

sed -E -e 's/.* ([^ ]+)$/\1/'

% echo 1 2 3 4 5 | sed -E -e 's/.* ([^ ]+)$/\1/'
5