MC AND 的解决方案有效,但它确实很慢(小测试样本需要约 1 秒)。
这是由于echo "!_buf!"|findstr ...
构造,对于每个字符,管道创建 cmd.exe 的两个实例并启动findstr
.
但这也可以通过纯批处理来解决。
测试每个字符是否在map
多变的
:test
set "_input=Th""i\s&& is not good _maybe_???"
set "_output="
set "map=abcdefghijklmnopqrstuvwxyz 1234567890"
:loop
if not defined _input goto endLoop
for /F "delims=*~ eol=*" %%C in ("!_input:~0,1!") do (
if "!map:%%C=!" NEQ "!map!" set "_output=!_output!%%C"
)
set "_input=!_input:~1!"
goto loop
:endLoop
echo(!_output!
当goto
循环被删除。
然后,您需要首先计算 stringLength,然后对每个字符使用 FOR/L 循环进行迭代。
该解决方案比上述方法快约 6 倍,比 MC ND 解决方案快约 40 倍
set "_input=Th""i\s&& is not good _maybe_!~*???"
set "_output="
set "map=abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ 1234567890"
%$strLen% len _input
for /L %%n in (0 1 %len%) DO (
for /F "delims=*~ eol=*" %%C in ("!_input:~%%n,1!") do (
if "!map:%%C=!" NEQ "!map!" set "_output=!_output!%%C"
)
)
exit /b
宏 $strlen 可以定义为
set LF=^
::Above 2 blank lines are required - do not remove
@set ^"\n=^^^%LF%%LF%^%LF%%LF%^^":::: StrLen pResult pString
set $strLen=for /L %%n in (1 1 2) do if %%n==2 (%\n%
for /F "tokens=1,2 delims=, " %%1 in ("!argv!") do (%\n%
set "str=A!%%~2!"%\n%
set "len=0"%\n%
for /l %%A in (12,-1,0) do (%\n%
set /a "len|=1<<%%A"%\n%
for %%B in (!len!) do if "!str:~%%B,1!"=="" set /a "len&=~1<<%%A"%\n%
)%\n%
for %%v in (!len!) do endlocal^&if "%%~b" neq "" (set "%%~1=%%v") else echo %%v%\n%
) %\n%
) ELSE setlocal enableDelayedExpansion ^& set argv=,