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Arronical
Asked: 2016-10-01 04:09:01 +0800 CST

Is there any way to make the cut command read the last field only?

  • 772

I have a space seperated string which is output by a command, which I like to pipe to cut, using -fd ' ' to split on spaces. I know I can use -f <n> to display field number <n>, but can I make it display the last field if I don't know the length of the string?

Or do I need to use a more flexible text editing tool like sed or awk?

command-line

4 Answers

  1. Best Answer

    No cut can't do that. You could use two rev commands, like

    echo 'foo bar baz' | rev | cut -d' ' -f1 | rev

    but it's usually easier to use awk:

    echo 'foo bar baz' | awk '{print $(NF)}'
    • 21

  2. You can do this using only shell, no external tool is needed, using Parameter Expansion:

    ${var##* }

    var##* will discard everything from start up to last space from parameter (variable) var.

    If the delimiter can be any whitespace e.g. space or tab, use character class [:blank:]:

    ${var##*[[:blank:]]}

    Example:

    $ var='foo bar spam egg'

$ echo "${var##* }"
egg


$ var=$'foo\tbar\tspam\tegg'

$ echo "$var"
foo    bar    spam    egg

$ echo "${var##*[[:blank:]]}"
egg
    • 10

  3. I personally like Florian Diesch answer. But there is this way too.

    a=$(echo "your string here" | wc -w)
echo "your string here" | cut -d' ' -f$a

    Explanation:

    wc -w gives the number of words. and cut cuts the last word

    EDIT:

    I figured another way of Doing it:

    echo "Any random string here" | tac -s' ' | head -1
    • 4

  4. Here is one using grep

    $ echo "Change is Good" | grep -o '[^ ]*$'
Good

    How it works:

    • grep with -o Print only the matched (non-empty) parts of a matching line.
    • The regexp [^ ]*$ matches anything from end until it found a space.

    Another one liner from glenn jackman using perl

    $ echo "Change is Good" | perl -lane 'print $F[-1]'
Good
    • 2