Linux结合sed和awk打印文件权限

问题: 过滤出stat /etc/hosts的权限值

法1:

[[email protected] ~]# stat /etc/hosts|sed -n ‘4p‘|awk -F ‘[0/]‘ ‘{print $2}‘

644

注:

Awk里的[0/]表示

Access: (0644/-rw-r--r--)

法2:

[[email protected] ~]# stat /etc/hosts|awk -F ‘[0/]‘ ‘NR==4 {print $2}‘

644

法3:

[[email protected] ~]# stat /etc/hosts|sed -n ‘4p‘

Access: (0644/-rw-r--r--) Uid: ( 0/ root) Gid: ( 0/ root)

stat /etc/hosts|sed -rn ‘4s###gp‘

stat /etc/hosts|sed -rn ‘4s#^.*##gp‘

Access: (0644/-rw-r--r--)

stat /etc/hosts|sed -rn ‘4s#^.*\(0##gp‘

Access: (0644/-rw-r--r--)

[[email protected] ~]# stat /etc/hosts|sed -nr ‘4s#^.*\(0(.*)/-.*$#\1#gp‘

644

Access: (0644/-rw-r--r--)