linux - 从配置文件中删除 ip 地址

Question

我有以下输出:

vif         = [ 'ip=1.2.3.4, mac=00:00:00:00:00:00, bridge=eth1', 'ip=5.6.7.8, mac=00:00:00:00:00:00, bridge=eth1' ]

有时候，只有一个ip地址。所以它是:

vif         = [ 'ip=1.2.3.4, mac=00:00:00:00:00:00, bridge=eth1' ]

而在其他情况下，有超过 2 个 ip 地址:

vif         = [ 'ip=1.2.3.4, mac=00:00:00:00:00:00, bridge=eth1', 'ip=5.6.7.8, mac=11:11:11:11:11:11, bridge=eth1', 'ip=9.1.2.3, mac=22:22:22:22:22:22, bridge=eth1' ]

有没有简单的方法只获取 ip 地址？我想将它们存储在一个数组中。

Answer 1

这是众多可能性中的一种:tr -s "[,'""\n"| grep "^ip="|剪切-d“=”-f2

例子:

echo "vif         = [ 'ip=1.2.3.4, mac=00:00:00:00:00:00, bridge=eth1', 'ip=5.6.7.8, mac=11:11:11:11:11:11, bridge=eth1', 'ip=9.1.2.3, mac=22:22:22:22:22:22, bridge=eth1' ]" | tr -s "[,'" "\n" | grep "^ip=" | cut -d "=" -f2

产生

1.2.3.4
5.6.7.8
9.1.2.3