Unix sort of version numbers

Question

I have a list of version numbers, let's say for instance that they are in a file versions.txt

1.2.100.4
1.2.3.4
10.1.2.3
9.1.2.3

I wish to sort them so that they are sorted by version. i.e:

1.2.3.4
1.2.100.4
9.1.2.3    
10.1.2.3

I have tried using various sort commands using the "k" parameters, but do not really understand it well enough to pull it off. Any help would be appreciated.

Answer 1

The -V option is the nicest, but I wanted to stay away from installing new/other software since my sort didn’t have that option.

This is the command that worked for me in the end:

sort -t. -k 1,1n -k 2,2n -k 3,3n -k 4,4n test.txt

From comments:

• To reverse the order: sort -t. -k 1,1nr -k 2,2nr -k 3,3nr -k 4,4nr
• To skip the v prefix: sort -t. -k 1.2,1n -k 2,2n -k 3,3n -k 4,4n

Answer 2

sort -V versions.txt

From man sort:

-V, --version-sort
natural sort of (version) numbers within text

See also Details about version sort.

Answer 3

BSD does not provide -V by default, so Ben's solution is as close as it gets. For your convenience I post here our version that is able to sort files like <label>-<version>.<ext>:

% ls bla-*.ime | sed -Ee 's/^(.*-)([0-9.]+)(\.ime)$/\2.-1 \1\2\3/'  | sort -t. -n -k1,1 -k2,2 -k3,3 -k4,4 | cut -d\  -f2-
bla-1.ime
bla-1.0.ime
bla-1.0.0.ime
bla-1.1.ime
bla-1.1.29.ime
bla-1.2.3.ime
bla-1.2.29.ime
bla-1.2.30.ime
bla-1.3.ime
bla-1.3.0.ime
bla-1.3.1.ime
bla-1.3.10.ime
bla-1.3.20.ime
bla-1.7.ime
bla-1.11.29.ime
bla-2.3.2.ime
bla-11.2.2.ime

Short explanation:

• List the files that you want to sort with ls.
• Find the version number and prefix the line with that.
• While doing that add -1 to the end to make shorter version number sort first (before .0 even). You could change -1 to 0 if you consider 1.3 to be equivalent to 1.3.0.
• Sort the lines using Ben's suggested solution on the version number.
• Chop off the version prefix from the line.

The list now contains a version sorted list of applicable file names. Any additional sorting on the label part is left as an exercise to the reader.

Answer 4

This command:

echo "1.2.100.4,1.2.3.4,10.1.2.3,9.1.2.3" | tr ',' '\n' | sort -V

Gives output:

1.2.3.4
1.2.100.4
9.1.2.3
10.1.2.3

Answer 5

In Perl:

sub compare_version_numbers {
   my ($l,$r) = @_;
   my @lx = split("\\.",$l);
   my @rx = split("\\.",$r);
   my $minlen = (@lx < @rx) ? @lx : @rx;
   for (my $i=0; $i < $minlen; $i++) {
      # make numeric by multiplying with 1
      my $l_number = ($lx[$i] * 1);
      my $r_number = ($rx[$i] * 1);
      # compare with spaceship operator
      my $l_vs_r = ($l_number <=> $r_number);
      # return if decision is clear!
      if ($l_vs_r != 0) {
         return $l_vs_r
      }
      # otherwise, next part in array of version numbers
   }
   # if we are here, we could not decide - shortest entry wins!
   return @lx <=> @rx
}

Answer 6

echo "1.2.100.4,1.2.3.4,10.1.2.3,9.1.2.3" | tr ',' '\n' | sort -k1,1n

Output:

1.2.100.4
1.2.3.4
9.1.2.3
10.1.2.3

You should be able to figure out the rest. Good luck

Answer 7

sort -n <versions.txt