How can I sort file names by version numbers?

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-se开发者_运维知识库tup

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?

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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.

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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.

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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

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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.

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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.

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$ 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

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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.