Format date and time in a Windows batch script

I ended up with this script:

set hour=%time:~0,2%
if "%hour:~0,1%" == " " set hour=0%hour:~1,1%
echo hour=%hour%
set min=%time:~3,2%
if "%min:~0,1%" == " " set min=0%min:~1,1%
echo min=%min%
set secs=%time:~6,2%
if "%secs:~0,1%" == " " set secs=0%secs:~1,1%
echo secs=%secs%

set year=%date:~-4%
echo year=%year%

:: On WIN2008R2 e.g. I needed to make your 'set month=%date:~3,2%' like below ::otherwise 00 appears for MONTH

set month=%date:~4,2%
if "%month:~0,1%" == " " set month=0%month:~1,1%
echo month=%month%
set day=%date:~0,2%
if "%day:~0,1%" == " " set day=0%day:~1,1%
echo day=%day%

set datetimef=%year%%month%%day%_%hour%%min%%secs%

echo datetimef=%datetimef%

I usually do it this way whenever I need a date/time string:

set dt=%DATE:~6,4%_%DATE:~3,2%_%DATE:~0,2%__%TIME:~0,2%_%TIME:~3,2%_%TIME:~6,2%
set dt=%dt: =0%

This is for the German date/time format (dd.mm.yyyy hh:mm:ss). Basically I concatenate the substrings and finally replace all spaces with zeros.

The resulting string has the format: yyyy_mm_dd__hh_mm_ss

---

Short explanation of how substrings work:

%VARIABLE:~num_chars_to_skip,num_chars_to_keep%

So to get just the year from a date like "29.03.2018" use:

%DATE:~6,4%
       ^-----skip 6 characters
         ^---keep 4 characters