Hello, Can anyone help with a BAT file which will open a txt file and delete the first four lines and the last five lines. The text file has a header and footer which needs to be removed: -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 NotDashEscaped: You need GnuPG to verify this message I have this code which deletes the first four lines (see below) does anyone have something I can add to it to delete the last 8 lines? Thanks in advance, Brett Code to delete the first 4 lines: @echo off setLocal EnableDelayedExpansion if exist *.new del *.new for /f "tokens=* delims= " %%A in ('dir /b *.txt') do ( set NAME=%%~nA set fileNAME=%%A for /f "tokens=* skip=4 delims= " %%L in (!fileNAME!) do ( echo %%L >> !NAME!.new ) ) ::==

This should work, BUT [here comes the but part] if there are blank lines it will take a bit more doing. ============================= @echo off > newfile & setLocal enableDELAYedexpansion set T= & set N= for /f "tokens=* delims= " %%a in (myfile) do ( set /a T+=1 ) set /a E=!T!-4 for /f "tokens=* delims= " %%a in (myfile) do ( set /a N+=1 if !N! gtr 4 ( if !N! lss !E! ( echo.%%a >> newfile ) ) ) ===================================== If at first you don't succeed, you're about average. M2