Getting a substring of a variable:

set a=abcdefgh
echo %a:~0,1%   & rem from index 0, length 1; result: a
echo %a:~1,1%   & rem from index 1, length 1; result: b
echo %a:~0,2%   & rem from index 0, length 2; result: ab
echo %a:~1,2%   & rem from index 1, length 2; result: bc
echo %a:~1%     & rem from index 1 to the end; result: bcdefgh
echo %a:~-1%    & rem from index -1 (last char) to the end; result: h
echo %a:~-2%    & rem from index -2 (next-to-last) to the end; result: gh
echo %a:~0,-2%  & rem from index 0 to index -2, excl.; result: abcdef
echo %a:~0,-1%  & rem from index 0 to index -1, excl.; result: abcdefg
echo %a:~1,-1%  & rem from index 1 to index -1, excl.; result: bcdefg

Testing substring containment:

Testing for "starts with":

if %a:~0,1%==a echo yes   & rem If variable a starts with "a", echo "yes".
if %a:~0,2%==ab echo yes  & rem If variable a starts with "ab", echo "yes".

String replacement:

set a=abcd & echo %a:c=%   & rem replace c with nothing; result: abd
set a=abcd & echo %a:c=e%  & rem replace c with e; result: abed; 
set a=abcd & echo %a:*c=%  & rem replace all up to c with nothing; result: d
rem Above, the asterisk (*) only works at the beginning of the sought pattern.

See also the help for SET command: set /?.

Splitting a string by any of " ", ",", and ";": ["space", "comma" and "semicolon":]

set myvar=a b,c;d
for %%a in (%myvar%) do echo %%a

Splitting a string by semicolon, assuming the string contains no quotation marks:

@echo off
set myvar=a b;c;d
set strippedvar=%myvar%
:repeat
for /f "delims=;" %%a in ("%strippedvar%") do echo %%a
set prestrippedvar=%strippedvar%
set strippedvar=%strippedvar:*;=%
if not "%prestrippedvar:;=%"=="%prestrippedvar%" goto :repeat

Limitations:

Links: