echo for /f "delims=" %%%i in ^(^'DIR /A -H /B^'^) do set "check^=^^!check^^!,"%%%i"" >> Uninstall.cmd
结果是
for /f "delims=" %%i""
我试图将原始代码回显到Uninstall.cmd
for /f "delims=" %%i in ('DIR /A -H /B') do set "check=!check!,"%%i""
谁能帮助我弄清楚如何使该行代码完全转义,以便可以将其输出到文件中?
如果您没有启用延迟扩展
echo for /f "delims=" %%%%i in ('DIR /A-H /B') do set "check=!check!,"%%%%i""
如果您启用了延迟扩展,则需要转义感叹号
echo for /f "delims=" %%%%i in ('DIR /A-H /B') do set "check=^!check^!,"%%%%i""
但是,如果代码是在一个块内执行的,则也需要转义右括号
if 1==1 (
echo for /f "delims=" %%%%i in ('DIR /A-H /B'^) do set "check=!check!,"%%%%i""
:: OR
echo for /f "delims=" %%%%i in ('DIR /A-H /B'^) do set "check=^!check^!,"%%%%i""
)
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我来说两句