China DOS Union

How can I extract lines 3 through 90 from a file and pass them to a new *.txt file! Thanks!

skip 2 lines, read once >new.txt

skip 89 lines, read once >>new.txt

I seem to have written one before, but I can't find it.

Could post #4 explain the code?

@echo off

for /f "skip=2 delims=" %%i in (1.txt) do (
@echo %%i>>Out.txt
goto Next
)
:Next
for /f "skip=89 delims=" %%i in (1.txt) do (
@echo %%i>>Out.txt
goto Exit
)
:Exit
pause>nul


I still think my own way is better... If there are a lot of lines, the FINDSTR above has to add line numbers to the whole text... and then filter...

If what the OP wants is the content from line 3 to line 90, then you can do it like this

@echo off
for /l %%a in (3,1,90) do call :need %%a
:need
for /f "tokens=1* delims=:" %%i in ('findstr /n . 1.txt') do if %%i==%1 echo %%j>>2.txt

Output in the original format, cannot output blank lines
@echo off for /l %%n in (3,1,90) do call :need %%n pause exit :need set /a line=%1-1 for /f "skip=%line% delims=" %%i in ('more test.txt') do ( echo %%i goto :eof )

Can output blank lines, but each line must not contain the “line number:” characters that are the same as the line number, and it cannot output the “!、:” characters or the characters enclosed by “!、:” on the same line
@echo off setlocal EnableDelayedExpansion for /l %%n in (31,90) do call :need %%n pause exit :need set /a line=%1-1 for /f "skip=%line% delims=" %%i in ('findstr /n .* "test.txt"') do ( set "str=%%i" set "str=!str:%1:=!" if not "!str!" == "" (echo !str!) else echo. goto :eof )

[ Last edited by zh159 on 2007-4-27 at 10:49 AM ]