在windows xp中使用批处理脚本(.bat文件),我将如何阅读文本文件并查找存在多少个字符实例?
例如,我有一个包含以下内容的字符串:
""OIJEFJ"JOIEJKAJF"""LKAJFKLJEIJ""JKLFJALKJF"LKJLKFA""""LKJKLFJLKADJF
我想要计算文件中有多少"
并返回计数。
答案 0 :(得分:8)
让我们开始计算一行中的字符。首先是缓慢而清晰的方法:
set i=-1
set n=0
:nextChar
set /A i+=1
set c=!theLine:~%i%,1!
if "!c!" == "" goto endLine
if !c! == !theChar! set /A n+=1
goto nextChar
:endLine
echo %n% chars found
现在快速而神秘的方法:
call :strLen "!theLine!"
set totalChars=%errorlevel%
set strippedLine=!theLine:%theChar%=!
call :strLen "!strippedLine!"
set /A n=totalChars-%errorlevel%
echo %n% chars found
goto :eof
:strLen
echo "%~1"> StrLen
for %%a in (StrLen) do set /A StrLen=%%~Za-4
exit /B %strLen%
最后计算文件中字符的方法:
set result=0
for /F "delims=" %%a in ('findstr "!theChar!" TheFile.txt') do (
set "theLine=%%a"
place the fast and cryptic method here
set /A result+=n
)
echo %result% chars found