In batch, how would I remove all non alphanumeric (a-z,A-Z,0-9,_) characters from a variable?
I\'m pretty sure I need to use findstr and a regex.
The solutionof MC ND works, but it's really slow (Needs ~1second for the small test sample).
This is caused by the echo "!_buf!"|findstr ...
construct, as for each character the pipe creates two instances of cmd.exe and starts findstr
.
But this can be solved also with pure batch.
Each character is tested if it is in the map
variable
: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!
And it could be speed up when the goto
loop is removed.
Then you need to calculate the stringLength first and iterate then with a FOR/L loop over each character.
This solution is ~6 times faster than the above method and ~40 times faster than the solution of MC ND
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
The macro $strlen can be defined with
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=,