bash - How can I sort file names by version numbers?

Question

In the directory "data" are these files:

command-1.9a-setup
command-2.0a-setup
command-2.0c-setup
command-2.0-setup

I would like to sort the files to get this result:

command-1.9a-setup
command-2.0-setup
command-2.0a-setup
command-2.0c-setup

I tried this

find /data/ -name 'command-*-setup' | sort --version-sort --field-separator=- -k2 

but the output was

command-1.9a-setup
command-2.0a-setup
command-2.0c-setup
command-2.0-setup

The only way I found that gave me my desired output was

tree -v /data

How could I get with sort the output in the wanted order?

Answer 1

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 /;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-setup
2.0.c command-2.0c-setup
2.0.a command-2.0a-setup
2.0 command-2.0-setup
10 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.