匹配末尾的端口号

Question

提取这个字符串的末尾端口号

2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215

我可以用grep awk 做出来

1.grep   echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| grep -Po  'd+$'
2.ack  echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| ack -o  'd+$'

为何下面的sed awk 无法做到？

3.sed    echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| sed  's/(.*)::(d+)$/2/g'
4.awk   echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| awk '{b=gensub(/^(.+)d+)$/,"2","g";print b}'

Answer 1

1. 在awk/sed中, d并不能代表数字, 数字只能用[0-9]来匹配, 所以d匹配都是失败的

2. awk中有语法错误: gensub括号不成对, 在引用捕获分组时, 需要 \, 不能单单一个, 因为shell层面也要转义一次

3. sed虽然没有语法错误, 但是匹配样式错了, 多了:
   修改后的答案:

awk:
    echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| awk '{print gensub(/^.+:([0-9]+)$/,"\1","g")}'
sed:
    echo "2017-04-17 08:16:14 INFO     connecting lh3.googleusercontent.com:443 from 111.111.111.111:26215"| sed -r  's/(.*):([0-9]+)$/2/g'