请教 sed 问题 - V2EX
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mrco
V2EX    Linux

请教 sed 问题

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  •   mrco Jun 10, 2015 4202 views
    This topic created in 3989 days ago, the information mentioned may be changed or developed.

    <mac address='52:54:00:a9:cc:20'/>

    sed如何过滤出52:54:00:a9:cc:20 ?

    sed "s/.'(.)'.*/\1/g" 这是可以的,但是不太明白具体含义,请赐教!

    Supplement 1    Jun 10, 2015
    上面sed命令有误,是这个:
    sed "s/.*'\(.*\)'.*/\1/g"
    8 replies    2015-06-11 07:59:11 +08:00
    lilydjwg
        1
    lilydjwg  
       Jun 10, 2015
    为什么不用 grep -o?
    knightdf
        2
    knightdf  
       Jun 10, 2015
    过滤为何不用grep -E ?
    还有: sed: 1: "s/.'(.)'.*/\1/g": \1 not defined in the RE 你确定这可以?
    sed: -e expression #1, char 15: invalid reference \1 on `s' command's RHS
    mrco
        3
    mrco  
    OP
       Jun 10, 2015
    @lilydjwg grep -o "..:.*:.." 可以实现,谢谢。
    但是还是想把sed这个搞明白
    kily
        4
    kily  
       Jun 10, 2015
    正确的应该是 sed "s/.*'\(.*\)'.*/\1/g" 吧
    你需要搜一下正则表达式的指南
    knightdf
        5
    knightdf  
       Jun 10, 2015
    @mrco 就是一个正则啊,对‘’里面的内容分组了然后后面\1填充捕获的第一个分组
    mrco
        6
    mrco  
    OP
       Jun 10, 2015
    @knightdf 诶,正则没学好。。。终于看明白了。
    .*'\(.*\)'.* 就是正则描述的整个字符串,然后是通过‘ ’分割的。
    然后用sed的第一个变量覆盖整个字符串。

    tks!
    lilydjwg
        7
    lilydjwg  
       Jun 11, 2015
    @mrco 嗯。所以你在主题帖里的代码根本就是错的,我还看了半天……
    lilydjwg
        8
    lilydjwg  
       Jun 11, 2015
    @mrco 这也是我反对在论坛等地默认使用 markdown 的原因。不知道的人写好了发出去,结果代码什么的都变了样。
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