string - 批处理 : String substitution sub

Question

我正在尝试批量制作一个子程序，在字符串中查找一个字符串并将其替换为第三个字符串。四处寻找，我发现here我可以使用 SET 命令删除字符串。

所以，这就是我尝试过的:

:modifyString what with [in]
SET _what=%~1
ECHO "%_what%"
SET "_with=%~2
ECHO "%toWith%"
SET _In=%~3
ECHO "%_In%"
SET _In=%_In:%toWhat%=%toWith%%
ECHO %_In%
SET "%~3=%_In%"
EXIT /B

ECHO 仅用于“调试”目的。

据我所知，错误在于...

SET _In=%_In:%toWhat%=%toWith%%

...因为 % 字符关闭了 %_in% 变量。

我也尝试过诸如...

SET _In=%_In:!toWhat!=!toWith!%
SET _In=!_In:%toWhat%=%toWith%!

...以及其他没有意义的人。

这是主要问题，另一个是在[in]中返回%_In%。

有什么建议吗？

Answer 1

这是使用 ! DelayedExpansion 方法的示例。

@echo off
setlocal EnableExtensions EnableDelayedExpansion
set "xxString=All your base are belong to us"
set "xxSubString=your base are belong"
set "xxNewSubString=of your bases belong"
echo Before
set xx
echo.
set "xxString=!xxString:%xxSubString%=%xxNewSubString%!"
echo After
set xx
endlocal
pause >nul

输出

Before
xxNewSubString=of your bases belong
xxString=All your base are belong to us
xxSubString=your base are belong

After
xxNewSubString=of your bases belong
xxString=All of your bases belong to us
xxSubString=your base are belong

修复你的

@echo off
:: Make sure that you have delayed expansion enabled.
setlocal EnableDelayedExpansion

:modifyString what with [in]
SET "_what=%~1"
SET "_with=%~2"
SET "_In=%~3"
ECHO "%_what%"
ECHO "%_with%"
ECHO "%_In%"
:: The variable names were not the same as the ones
:: defined above.
SET _In=!_In:%_what%=%_with%!
ECHO %_In%

:: This will not change the value of the 3rd parameter
:: but instead will create a new parameter with the
:: value of %3 as the variable name.
SET "%~3=%_In%"
endlocal
EXIT /B

如何在不延迟扩展的情况下进行子字符串替换。使用call命令创建两级变量扩展。在变量周围使用单个 % 来扩展第一个变量，并在变量周围使用两个 %% 来扩展第二个。

@echo off
setlocal EnableExtensions
set "xxString=All your base are belong to us"
set "xxSubString=your base are belong"
set "xxNewSubString=of your bases belong"
echo Before
set xx
echo.
call set "xxString=%%xxString:%xxSubString%=%xxNewSubString%%%"
echo After
set xx
endlocal
pause >nul