How to print only last column?

Question

echo -e 'one two three\nfour five six\nseven eight nine'
one two three
four five six
seven eight nine

how can I do some "MAGIC" do get this output?:

three
six
nine

UPDATE: I don't need it in this specific way, I need a general solution so that no matter how many columns are in a row, e.g.: awk always displays the last column.

Answer 1

It can even be done only with 'bash', without 'sed', 'awk' or 'perl':

echo -e 'one two three\nfour five six\nseven eight nine' |
  while IFS=" " read -r -a line; do
    nb=${#line[@]}
    echo ${line[$((nb - 1))]}
  done

Answer 2

echo -e 'one two three\nfour five six\nseven eight nine' | awk '{print $3}'

EDIT:

echo -e 'one two three\nfour five six\nseven eight nine' | awk '{print $NF}'

Answer 3

It can also be done using 'sed':

echo -e 'one two three\nfour five six\nseven eight nine' | sed -e 's/^.* \([^ ]*\)$/\1/'

Update:

or more simply:

echo -e 'one two three\nfour five six\nseven eight nine' | sed -e 's/^.* //'

Answer 4

It's easier than you think.

$ echo one two three | awk '{print $NF}'
three

Answer 5

... | perl -lane 'print $F[-1]'

Answer 6

Or using cut:

echo -e 'one two three\nfour five six\nseven eight nine' | cut -f 3 -d' '

although this does not satisfy the 'general solution' requirement. Using rev twice we can solve this as well:

echo -e 'one two three\nfour five six\nseven eight nine' | rev | cut -f 1 -d' ' | rev

Answer 7

Using awk you can first check if there is at least one column.

echo | awk '{if (NF >= 1) print $NF}'

echo 1 2 3 | awk '{if (NF >= 1) print $NF}'