I need to count the characters of an inputed string in Batch. I don't want to use temporary files. Could it be done without them? If yes, explanations of your code would be greatly appreciated. Thanks SO!
A simple way is to use a function
@echo off
set "myVar=abcdefg"
call :Stringlength result myVar
echo %result%
exit /b
:Stringlength <resultVar> <stringVar>
(
setlocal EnableDelayedExpansion
set "s=!%~2!#"
set "len=0"
for %%P in (4096 2048 1024 512 256 128 64 32 16 8 4 2 1) do (
if "!s:~%%P,1!" NEQ "" (
set /a "len+=%%P"
set "s=!s:~%%P!"
)
)
)
(
endlocal
set "%~1=%len%"
exit /b
)
This can measure the string to a maximum of 8192 characters, as the maximum size of a string is 8191 bytes, this should be enough.
The first parenthesis blocks is only for a bit more performance.
The second block is needed to return the %len% value behind the endlocal barrier.
The main idea is a binary search, in the first loop the temporary copy in s of the string is tested if it is longer than 4096 bytes or not.
Then the next test will be with 2048 or 6144 (=2048+4096), so the len variable will be at each loop a little bit more exact.
After 13 loops the len is exact.
For faster strlen functions you could read strlen boosted, which uses some more tricks.
There is also a solution with batch macros, macros are normally much faster than functions in batch.
@echo off
call :loadMacros
set "myVar=abcdefg"
%$strlen% result myVar
echo %result%
exit /b
:loadMacros
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=,
exit /b
At dostips.com are some discussion about the macro technic
1 Batch "macros" with arguments
2 macros with appended parameters
When you call the function, the 2nd parameter should be a value rather than a reference:
call :Stringlength result %myVar%
来源：https://stackoverflow.com/questions/10672885/how-to-count-the-characters-in-a-string-with-batch