如何获得这个输出？perl/awk/sed？如何？

Question

我如何在“一般”中从这个 INPUT 生成

输入 (/proc/net/ip_conntrack)

udp      17 0 src=192.168.1.128 dst=91.120.112.125 sport=29249 dport=39802 packets=3 bytes=408 [UNREPLIED] src=91.120.112.125 dst=79.132.235.112 sport=39802 dport=29249 packets=0 bytes=0 mark=0 use=2
udp      17 146 src=192.168.1.128 dst=98.196.37.3 sport=56932 dport=43645 packets=924 bytes=406167 src=98.196.37.3 dst=79.132.235.112 sport=43645 dport=56932 packets=1042 bytes=546092 [ASSURED] mark=0 use=2
tcp      6 118 SYN_SENT src=192.168.1.129 dst=89.132.51.110 sport=2518 dport=47385 packets=2 bytes=104 [UNREPLIED] src=80.132.51.190 dst=79.132.235.112 sport=47385 dport=2518 packets=0 bytes=0 mark=0 use=2
udp      17 147 src=192.168.1.128 dst=98.196.37.7 sport=56937 dport=43647 packets=924 bytes=406167 src=98.196.37.7 dst=80.132.235.117 sport=43647 dport=56937 packets=1042 bytes=546092 [ASSURED] mark=0 use=2
tcp      6 119 SYN_SENT src=192.168.1.129 dst=89.132.51.110 sport=2514 dport=47384 packets=2 bytes=104 [UNREPLIED] src=80.132.51.194 dst=80.132.235.114 sport=47384 dport=2514 packets=0 bytes=0 mark=0 use=2
udp      17 163 src=192.168.1.111 dst=192.168.1.201 sport=67 dport=68 packets=29 bytes=9512 src=192.168.1.201 dst=192.168.1.1 sport=68 dport=67 packets=27 bytes=8856 [ASSURED] mark=0 use=2
tcp      6 1 TIME_WAIT src=192.168.1.201 dst=67.201.31.15 sport=55479 dport=80 packets=7 bytes=725 src=67.201.31.35 dst=79.132.235.112 sport=80 dport=55479 packets=5 bytes=1963 [ASSURED] mark=0 use=2

到这个输出？

输出

udp 192.168.1.128   3
tcp 192.168.1.129   2
udp 192.168.1.111   1
tcp 192.168.1.201   1

因此，从 INPUT 数据中，OUTPUT 将是 3 列：

protocoll   src-ip-address  count-connection-numbers

这样连接数最多的 IP 将位于顶部。

Answer 1

使用sed、uniq、sort、awk和column：

sed -n -r 's/^(udp|tcp)\s*[^=]*=([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*$/\1 \2/p' infile \
  | sort | uniq -c | sort -nr | awk '{print $2" "$3" "$1'} | column -t > outfile