sorting - How to sort lines of a text file containing version numbers in format major.minor.build.revision numerical?

Question

I have a txt file with values like this:

3.6.4.2
3.6.5.1
3.6.5.10
3.6.5.11
3.6.5.12
3.6.5.13
3.6.5.2
3.6.7.1
3.6.7.10
3.6.7.11
3.6.7.2
3.6.7.3

I need to write a batch script and return a sorted output. The problem is with last column, numbers .10 and .11 should go after .3 and so. I need the "latest version" to be on the bottom, which in this case is 3.6.7.11

In Linux I used "sort -t"." -k1n,1 -k2n,2 -k3n,3 -k4n,4" but I can't get it working with batch script.

Also I am not allowed to use Cygwin or PowerShell for some reasons.

In my batch code I am so far trying only various versions of this but nothing is working for me:

sort /+n versions.txt

The output used in this question is simply

sort versions.txt

It looks like that command sort is doing it correctly until I don't have 2 digits number used.

Answer 1

This is a common problem in Batch files. All sorting methods use a string comparison, where "10" comes before "2", so it is necessary to insert left zeros in the numbers less than 10. The Batch file below do that, but instead of generate a new file with the fixed numbers, it uses they to create an array that will be automatically sorted. After that, the array elements are shown in its natural (sorted) order.

EDIT: I modified the code in order to manage two digits numbers in the four parts.

@echo off
setlocal EnableDelayedExpansion

for /F "tokens=1-4 delims=." %%a in (input.txt) do (
    rem Patch the four numbers as a two digits ones
    set /A "a=100+%%a, b=100+%%b, c=100+%%c, d=100+%%d"
    rem Store line in the proper array element
    set "line[!a:~1!!b:~1!!c:~1!!d:~1!]=%%a.%%b.%%c.%%d"
)

rem Show array elements
for /F "tokens=2 delims==" %%a in ('set line[') do echo %%a

Output:

3.6.4.2
3.6.5.1
3.6.5.2
3.6.5.10
3.6.5.11
3.6.5.12
3.6.5.13
3.6.7.1
3.6.7.2
3.6.7.3
3.6.7.10
3.6.7.11