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

Answer 1

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.

Answer 2

Another way to do this is to pad your numbers.

This example pads all numbers to 8 digits. Then, it does a plain alphanumeric sort. Then, it removes the pad.

$ pad() { perl -pe 's/(\d+)/0000000\1/g' | perl -pe 's/0*(\d{8})/\1/g'; }
$ unpad() { perl -pe 's/0*([1-9]\d*|0)/\1/g'; }
$ cat files | pad | sort | unpad
command-1.9a-setup
command-2.0-setup
command-2.0a-setup
command-2.0c-setup
command-10.1-setup

To get some insight into how this works, let's look at the padded sorted result:

$ cat files | pad | sort
command-00000001.00000009a-setup
command-00000002.00000000-setup
command-00000002.00000000a-setup
command-00000002.00000000c-setup
command-00000010.00000001-setup

You'll see that with all the numbers nicely padded to 8 digits, the alphanumeric sort puts the filenames into their desired order.

Answer 3

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-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.

Answer 4

$ cat files
command-1.9a-setup
command-2.0c-setup
command-10.1-setup
command-2.0a-setup
command-2.0-setup

$ cat files | sort -t- -k2,2 -n
command-1.9a-setup
command-2.0-setup
command-2.0a-setup
command-2.0c-setup
command-10.1-setup

$ tac files | sort -t- -k2,2 -n
command-1.9a-setup
command-2.0-setup
command-2.0a-setup
command-2.0c-setup
command-10.1-setup

Answer 5

I have files in a folder and need to get those name in sort order, based on the number. E.g. -

abc_dr-1.txt
hg_io-5.txt
kls_er_we-3.txt
sd-4.txt
sl_rt_we_yh-2.txt

I need to sort them based on number. So I used this to sort.

ls -1 | sort -t '-' -nk2

It gave me files in sort order based on number.

Answer 6

Old post, but... ls -l --sort=version may be of assistance (although for OP's example the sort is the same as done by ls -l in a RHEL 7.2):

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

YMMV i guess.