Question

我想在文件中搜索数字64后添加新行。我喜欢使用awk或sed来做这件事。

64 bytes from 170.198.42.128: icmp_seq=1 ttl=60 time=83.4 ms 64 bytes from
170.198.42.128: icmp_seq=2 ttl=60 time=76.6 ms 64 bytes from 170.198.42.128: icmp_seq=3  
ttl=60 time=70.8 ms 64 bytes from 170.198.42.128: icmp_seq=4 ttl=60 time=76.6 ms --- 
170.198.42.128 ping statistics --- 4 packets transmitted, 4 received, 0% packet loss, time 
3000ms rtt min/avg/max/mdev = 70.861/76.924/83.493/4.473 ms

我使用此命令

cat file | sed 's/64/\n/g'

但这取代了数字64.我想要打破模式并显示以64开头正确的ping命令模式

我尝试使用追加和插入模式..但没有正确使用它们

需要帮助

Answer 1

在替换中，替换中的&将替换为匹配的任何内容。所以：

sed 's/64/\n&/g' file

Answer 2

这个sed命令应该适用于gnu和BSD sed版本：

sed $'s/64/\\\n&/g' file

64 bytes from 170.198.42.128: icmp_seq=1 ttl=60 time=83.4 ms
64 bytes from
170.198.42.128: icmp_seq=2 ttl=60 time=76.6 ms
64 bytes from 170.198.42.128: icmp_seq=3
ttl=60 time=70.8 ms
64 bytes from 170.198.42.128: icmp_seq=4 ttl=60 time=76.6 ms ---
170.198.42.128 ping statistics --- 4 packets transmitted, 4 received, 0% packet loss, time
3000ms rtt min/avg/max/mdev = 70.861/76.924/83.493/4.473 ms

更新：以下是gnu awk命令：

awk '1' RS='64' ORS='\n64' file

Answer 3

这可能更像你真正需要的（使用GNU awk进行多字符RS）：

$ gawk -v RS='^$' '{gsub(/[[:space:]]+/," "); gsub(/(\<[[:digit:]]+\> bytes|---)/,"\n&"); sub(/^\n/,"")}1' file
64 bytes from 170.198.42.128: icmp_seq=1 ttl=60 time=83.4 ms
64 bytes from 170.198.42.128: icmp_seq=2 ttl=60 time=76.6 ms
64 bytes from 170.198.42.128: icmp_seq=3 ttl=60 time=70.8 ms
64 bytes from 170.198.42.128: icmp_seq=4 ttl=60 time=76.6 ms
--- 170.198.42.128 ping statistics
--- 4 packets transmitted, 4 received, 0% packet loss, time 3000ms rtt min/avg/max/mdev = 70.861/76.924/83.493/4.473 ms

找到匹配后添加新行

3 个答案: