如何取得/etc/hosts文件的权限对应的数字内容，如-rw-r--r--为644，要求使用命令取得644这样的数字。
解答：
[[email protected] ~]# stat /etc/hosts
File: `/etc/hosts‘
Size: 216 Blocks:8 IO Block: 4096 regular file
Device: 803h/2051d Inode:260126 Links: 2
Access: (0644/-rw-r--r--) Uid: ( 0/ root) Gid: ( 0/ root)
Access: 2017-07-30 00:02:05.983001919 +0800
Modify: 2017-05-20 22:57:33.530934887 +0800
Change: 2017-05-20 22:57:33.533934704 +0800
[[email protected] ~]# stat /etc/hosts|awk ‘NR==4‘
Access: (0644/-rw-r--r--) Uid: ( 0/ root) Gid: ( 0/ root)
[[email protected] ~]# stat /etc/hosts|awk ‘NR==4‘|sed‘s#^.*(0##g‘
644/-rw-r--r--) Uid: ( 0/ root) Gid: ( 0/ root)
方法一：sed去掉前面，去掉后面
[[email protected] ~]# stat /etc/hosts|awk ‘NR==4‘|sed ‘s#^.*(0##g‘|sed‘s#/.*$##g‘
644
方法二：sed反向引用
[[email protected] ~]# stat /etc/hosts | sed -n ‘4p‘ | sed -r‘s#^.*\((.*)/-.*$#\1#g‘
0644
[[email protected] ~]#
[[email protected] ~]# stat /etc/hosts | sed -n ‘4p‘ | sed -r‘s#^.*\(0(.*)/-.*$#\1#g‘
644
方法三：awk分隔用法菜刀系列
[[email protected] ~]# stat /etc/hosts | awk ‘NR==4‘ |awk -F "[0/]"‘{print $2}‘
644
[[email protected] ~]#
[[email protected] ~]# stat /etc/hosts | awk ‘NR==4‘ |awk -F "[0/]"‘{print $2}‘
644
[[email protected] ~]#
[[email protected] ~]# stat /etc/hosts | awk -F "[0/]" ‘NR==4{print$2}‘
644
[[email protected] ~]#
[[email protected] ~]# stat /etc/hosts | awk -F "[(/]" ‘NR==4{print$2}‘
0644
4.命令的结果里面有你想要的东西----可能会有现成参数
# stat-c%a /etc/hosts
644