给出以下记录
01-01-2012 18:02 some data 01-11-2014 20:22 some other data 10-02-2014 14:00 more data still
我正在尝试将日期,时间和数据进行分组,并将它们打印在单独的行上,如下所示:
01-01-2012 18:02 some data
01-11-2014 20:22 some other data
10-02-2014 14:00 more data still
但是,到目前为止,我有:
echo '01-01-2012 18:02 some data 01-11-2014 20:22 some other data 10-02-2014 14:00 more data still' | awk -F '[0-9]*-[0-9]*-[0-9]* [0-9]*:[0-9]*' '{ for ( n=1; n<=NF; n++ ) print $n }
结果是:
some data
some other data
more data still
日期和时间丢失。它们是字段分隔符,因此不打印。
我如何修改我的awk脚本以打印与正则表达式匹配的每个字段分隔符?
使用gnu awk:
awk -v RS='[0-9]+-[0-9]+-[0-9]+ [0-9]+:[0-9]+' '!NF{s=RT;next} {print s $0}' file
01-01-2012 18:02 some data
01-01-2012 18:02 some other data
01-01-2012 18:02 more data still
编辑:使用非gnu awk,您可以执行以下操作:
awk '{gsub(/[[:blank:]]+[0-9]+-[0-9]+-[0-9]+ [0-9]+:[0-9]+/, "\n&");
gsub(/\n[[:blank:]]+/, "\n")} 1' file
01-01-2012 18:02 some data
01-11-2014 20:22 some other data
10-02-2014 14:00 more data still
也grep -P
可以使用:
grep -oP '[0-9]+-[0-9]+-[0-9]+ [0-9]+:[0-9]+.+?(?=[0-9]+-[0-9]+-[0-9]+|$)' file
01-01-2012 18:02 some data
01-11-2014 20:22 some other data
10-02-2014 14:00 more data still
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