How can I sort file names by version numbers?

Edit: It turns out that Benoit was sort of on the right track and Roland tipped the balance

You simply need to tell sort to consider only field 2 (add ",2"):

find ... | sort --version-sort --field-separator=- --key=2,2

Original Answer: ignore

If none of your filenames contain spaces between the hyphens, you can try this:

find ... | sed 's/.*-\([^-]*\)-.*/\1 \0/;s/[^0-9] /.&/' | sort --version-sort --field-separator=- --key=2 | sed 's/[^ ]* //'

The first sed command makes the lines look like this (I added "10" to show that the sort is numeric):

1.9.a command-1.9a-setup2.0.c command-2.0c-setup2.0.a command-2.0a-setup2.0 command-2.0-setup10 command-10-setup

The extra dot makes the letter suffixed version number sort after the version number without the suffix. The second sed command removes the prefixed version number from each line.

There are lots of ways this can fail.

If you specify to sort that you only want to consider the second field (-k2) don't complain that it does not consider the third one.

In your case, run sort --version-sort without any other argument, maybe this will suit better.

Looks like this works:

find /data/ -name 'command-*-setup' | sort -t - -V -k 2,2

not with sort but it works:

tree -ivL 1 /data/ | perl -nlE 'say if /\Acommand-[0-9][0-9a-z.]*-setup\z/'

-v: sort the output by version
-i: makes tree not print the indentation lines
-L level: max display depth of the directory tree