将批处理文件中的特殊字符替换为%

Question

任何人都能帮助如何替换包含空格、（、-、）等特殊字符的字符串吗。假设我有一个字符串：

str=A Room Has (1 level) - (2 - leve) Rooms
Output string should be like:
str=A%Room%Has%%1%level%%%%2%%%leve%%Rooms

Answer 1

这应该有效：

set "str=A Room Has (1 level) - (2 - leve) Rooms"
set "str=!str:-=%%!"
set "str=!str:(=%%!"
set "str=!str:)=%%!"
set "str=!str: =%%!"

Answer 2

这样尝试：

@echo off
set "str=A Room Has (1 level) - (2 - leve) Rooms"
echo Before string substitution :   %str%
pause
setlocal EnableDelayedExpansion
set "str=!str:-=%%!"
set "str=!str:(=%%!"
set "str=!str:)=%%!"
set "str=!str: =%%!"
Color 0A
echo After string substitution  :   !str!
pause