Question

用批处理提取指定路径下所有.txt文件的前N行，合并到指定文件

例如，提取 D:\aam\ 下所有.txt文件的前10行，合并输出到 D:\axs\add.txt

Answer 1

@echo off
setlocal enabledelayedexpansion
cd /d D:\aam\
for /f "tokens=*" %%i in ('dir/b D:\aam\*.txt') do (
set n=0
for /f "tokens=*" %%j in ('type "%%i"') do (
if !n! lss 10 echo %%j>>D:\axs\add.txt
set /a n+=1
)
)
endlocal

=====================================================================
灵活高效：

@echo off
setlocal enabledelayedexpansion

set 源目录=D:\aam\
set 目标文件=D:\axs\add.txt
set 截止行=10

if "%源目录:~-1,1%"=="\" set "源目录=%源目录:~,-1%"
cd /d "%源目录%"
cd.>"%目标文件%"
for /f "tokens=*" %%i in ('dir/b *.txt') do call :xwj "%源目录%\%%i"
endlocal
goto :eof
:xwj
set n=0
for /f "tokens=*" %%j in ('type "%~1"') do (
set /a n+=1
echo %%j>>"%目标文件%"
if !n!==%截止行% goto :eof
)
goto :eof

Answer 2

@echo off
setlocal enabledelayedexpansion
for /f %%i in ('dir/b d:\aam\*.txt‘）do (
set /a n=0
for /f %%j in ("%%i") do (
if !n! lss 10 echo %%j >>d:\axs\add.txt
set /a n+=1
)
)