sed和awk关于小括号里面的匹配

Linux9253

1. [stock@localhost ~]$ id root
   
2. uid=0(root) gid=0(root) groups=0(root),1(bin),2(daemon),3(sys),4(adm),6(disk),10(wheel)
   
3. [stock@localhost ~]$

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要求用sed和awk匹配出小口号里面的内容
root toot root bin daemon sys adm disk wheel
求解、、、、、

winway1988

1. sed 'G;:a;s/(\([^)]*\))\(.*\)/\2 \1/;ta;s/.*\n //'

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winway1988

1. awk -vRS="\\\([^)]*\\\)" -vORS=" " 'RT{gsub(/[()]/,"",RT);print RT}'

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hbmhalley

1. sed 's/\(^\|)\)[^(]*\((\|$\)/ /g'

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huohulwz

1. sed 's/[^()]*(//g; y#)# #'

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dn833

1. awk -F "[()]" '{for(i=2;i<=NF;i+=2){print $i}}'

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如果都是这么有规律成对的（）就可以了

jason680

回复 1# Linux9253

How about this one

# id -nG root
root bin daemon sys adm disk wheel

# echo `id -nu root` `id -ng root` `id -nG root`
root root root bin daemon sys adm disk wheel
   

07101060407

1. id root|sed  's/[^)(]*(\([^)(]*\)[^)(]*)/\1 /g'
   
2. id root|sed  's/[^)(]*(\|)/ /g'

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07101060407

1. id root|grep -ioP '(?<=\().*?(?=\))'

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