Batch script to find and replace a string in text file within a minute for files up to 12 MB

Question

I have written a batch script to replace a string in text file.

Following is the script.

@echo off &setlocal set "search=%1" set "replace=%2" set "textfile=Input.txt" set "newfile=Output.txt" (for /f "delims=" %%i in (%textfile%) do (     set "line=%%i"     setlocal enabledelayedexpansion     set "line=!line:%search%=%replace%!"     echo(!line!     endlocal ))>"%newfile%" del %textfile% rename %newfile%  %textfile% 

But for a 12MB file, it is taking close to 7 min. I want it to be under a minute. Can we make use of find or findstr command to reduce the time taken?

Answer 1

Give this a shot:

@echo off setlocal  call :FindReplace "findstr" "replacestr" input.txt  exit /b   :FindReplace <findstr> <replstr> <file> set tmp="%temp%\tmp.txt" If not exist %temp%\_.vbs call :MakeReplace for /f "tokens=*" %%a in ('dir "%3" /s /b /a-d /on') do (   for /f "usebackq" %%b in (`Findstr /mic:"%~1" "%%a"`) do (     echo(&Echo Replacing "%~1" with "%~2" in file %%~nxa     <%%a cscript //nologo %temp%\_.vbs "%~1" "%~2">%tmp%     if exist %tmp% move /Y %tmp% "%%~dpnxa">nul   ) ) del %temp%\_.vbs exit /b  :MakeReplace >%temp%\_.vbs echo with Wscript >>%temp%\_.vbs echo set args=.arguments >>%temp%\_.vbs echo .StdOut.Write _ >>%temp%\_.vbs echo Replace(.StdIn.ReadAll,args(0),args(1),1,-1,1) >>%temp%\_.vbs echo end with