Print the last field using 'cut'

Question

Let's say I've the following data:

ae722d1d94dcd3b161af166936:!]z@.:123
09338b2b29919e6c0daefbcce0361335:21f2e48d86f28ab7cd5c9c95:123
$H$92Gh.gRJwECLPzkjACPihoLg/:123
c6f1cb5f1feeece60f9a8e067e0:v4Zz'Cj5_Ze{J+iRW{2z,<~:123
202cb962a75b964b07152d234b70:123
40bd0015630c35165329ea5ecbdbbeef:123

(Invaild hashes, for explaining purpose only)

I want to use the cut tool in printing the last field only.

123
123
123
123
123
123

• I don't want to use rev command.
• I want nothing other than cut, I know how to do it in sed, awk.

--complement flag might help!

Answer 1

For that specific data you can:

cut -f2- -d: file | cut -f2 -d:

If you might have three colons:

cut -f2- -d: file | cut -f2- -d: | cut -f2 -d:

You can keep adding more cuts as needed.

The trick is that cut does nothing to a record that doesn't have the delimiter so you can use successive cuts to chop off the first field while leaving the last fields that you've already found alone.

Answer 2

i saw this on another place when I was looking for this answer.

rev, cut (on delimiter) -f 1, rev

move the last field forward, cut it, and move it back

i thought that was kinda slick