[文本处理] gawk怎么删除txt的注释

mio

怎么用gawk删掉txt里面！或者#开头的行

Batcher

回复 1# mio

gawk "!/^[!#]/" a.txt > b.txt
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cmd1152

刚刚忘记感叹号了...

@echo off
set file=文件.txt
(FOR /F "eol=# delims=" %%i in (%file%) do (echo %%i))>%file%.tmp
(FOR /F "eol=! delims=" %%i in (%file%.tmp) do (echo %%i))>%file%.temp
type %file%.temp>%file%
del %file%.temp %file%.tmp
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idwma

没有gawk正则的写法应该大同小异
findstr /v "^# ^!" 1.txt

newswan

sed

sed -r -e "s/^[!#].*//" -i a.txt
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sed -r -e "/^[!#]/d" -i a.txt
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