Portál AbcLinuxu, 7. května 2025 01:12
Řešení dotazu:
subnet 192.168.1.0 netmask 255.255.255.0 {
range 192.168.1.100 192.168.1.199;
option broadcast-address 192.168.1.255;
option subnet-mask 255.255.255.0;
option routers 192.168.1.1;
}
subnet 192.168.2.0 netmask 255.255.255.0 {
range 192.168.2.100 192.168.2.199;
option broadcast-address 192.168.2.255;
option subnet-mask 255.255.255.0;
option routers 192.168.2.1;
}
host my-host-1 {
hardware ethernet 01:02:03:04:05:06;
fixed-address 192.168.1.10;
}
host my-host-2 {
hardware ethernet 01:02:03:04:05:06;
fixed-address 192.168.2.10;
}
subnet 192.168.1.0 netmask 255.255.255.0 {
range 192.168.1.100 192.168.1.199;
option broadcast-address 192.168.1.255;
option subnet-mask 255.255.255.0;
option routers 192.168.1.1;
}
subnet 192.168.2.0 netmask 255.255.255.0 {
range 192.168.2.100 192.168.2.199;
option broadcast-address 192.168.2.255;
option subnet-mask 255.255.255.0;
option routers 192.168.2.1;
}
subnet 192.168.3.0 netmask 255.255.255.0 {
range 192.168.2.100 192.168.3.199;
option broadcast-address 192.168.3.255;
option subnet-mask 255.255.255.0;
option routers 192.168.3.1;
}
...
host my-host-1 {
hardware ethernet 01:02:03:04:05:06;
fixed-address 192.168.1.10;
}
host my-host-2 {
hardware ethernet 01:02:03:04:05:06;
}
Takze host s MAC 01:02:03:04:05:06 v subnetu 192.168.1.0 dostane 192.168.1.10 adresu, ve vsech ostatnich by mel mit dynamickou adresu.
class "static" { match if binary-to-ascii (16,8,":",substring(hardware, 0, 4)) = "1:0:60:60:11:22:33:44"; }a potom aplikovat class na jednotlive subnety s tim, ze v druhem pripade zadefinujes smostatny pool s jednou IP adresou. NN
Tiskni
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