How can I generate from this INPUT in "general"
INPUT (/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=192.168.1.1 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
to this output?
OUTPUT
udp 192.168.1.128 3
tcp 192.168.1.129 2
udp 192.168.1.1 1
tcp 192.168.1.201 1
So that from the INPUT datas the OUTPUT would be 3 columns:
protocoll src-ip-address count-connection-numbers
So that the IP with the most number of connections would be at top.
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