Batch - replacing with percent symbol

Question

I want to replace "mod" in string with "%":
set string=%string:mod=x%
What I should input as "x"?

Answer 1

you can do that by enabling delayed expansion so you can use ! as delimiters. Then, doubling the percent sign allows to represent percent as a replacement char.

@echo off
setlocal enabledelayedexpansion

set string=12 mod 15
set string=!string:mod=%%!
echo %string%

result

12 % 15