awk

回复 7# hankerer


但是，行首本来就没有空白字符啊。

hankerer

回复 5# awk


   前两行是去行首的空白字符的

awk

sed -r "s/ +/,/g" a.txt >b.txt
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awk

回复 4# hankerer


保留最后一行，前面两行可以删掉吧？

hankerer

借一下fr工具

@echo off
for /f "tokens=*" %%i in (a.txt) do echo %%i>>b.txt
fr /r:" +" /t:"," b.txt
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awk

gawk -v OFS=, "$1=$1" a.txt >b.txt
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CrLf

@(for /f "delims=" %%a in (a.txt)do echo;&for %%b in (%%a)do set/p=%%b)<nul>b.txt
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%1 @cmd /q /v /c"for /f "delims^=" %%a in (a.txt)do set str=%%a&(for /l %%b in (-9 1)do set str=!str:  = !)&echo !str: =,!"
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手机不便，怎么短怎么写，不考虑可读性…