What is missing in forwarding between interfaces?

Question

css

For example, there are 4 interfaces:
eth0, eth1, eth2, eth3
Internet comes to eth0, which must be "distributed" to eth1
. Another Internet comes to eth2, which must be "distributed" to eth3. Networks should not overlap.
I do like this:

eth0: static 10.100.7.154/16 gw 10.100.0.1 
eth2: static 10.100.7.155/16 gw 10.100.0.1

eth1: 10.0.1.1/24+dhcp_server
eth2 :10.0.2.1/24+dhcp_server

iptables:

iptables -A FORWARD -s 10.0.1.0/24 -i eth1 -o eth0 -j ACCEPT
iptables -A FORWARD -d 10.0.1.0/24 -i eth0 -o eth1 -j ACCEPT
iptables -A FORWARD -s 10.0.2.0/24 -i eth3 -o eth2 -j ACCEPT
iptables -A FORWARD -d 10.0.2.0/24 -i eth2 -o eth3 -j ACCEPT

NAT:

iptables -t nat - a POSTROUTING -s 10.0.1.0/24 -o eth0 -j SNAT --to-source 10.100.7.154
iptables -t nat - a POSTROUTING -s 10.0.2.0/24 -o eth2 -j SNAT --to-source 10.100.7.155

Question:
why is there Internet on the eth1 interface, but not on eth3. Everything seems to be done the same way.

Answer 1

It is not clear what is wrong with the fact that there are many, to list the problem with a comma. And so it would be possible to see html to understand.
For the for to work, as far as I remember, the id of the checkboxes will have to be specified, and if you remove the id, then the label will not be sent to the checkbox. And then the only thing you can do is

.pay:checked ~ [for="payt4"]:before {
    background:#d9d9d9;
    color: #333;
}

and since there are a lot of them

.pay:checked ~ [for="payt3"]:before ,
.pay:checked ~ [for="payt4"]:before {
    background:#d9d9d9;
    color: #333;
}

you can take a chance and try

.pay:checked ~ label:before, {
    background:#d9d9d9;
    color: #333;
}

or

.pay:checked ~ label[for]:before, {
    background:#d9d9d9;
    color: #333;
}

Only each .pay and label will need some kind of block so that they do not affect each other or subsequent blocks

Answer 2

you can wrap each input in a label, then you don't need for="payt4"

Answer 3

Example

Answer 4

you have eth0 and eth2 ip addresses from the same subnet /16. This not normal.

Answer 5

ip rule to help you. Along with additional routing tables