Welcome to the Linux Foundation Forum!

10 - Block attacker IP with express

chesterhengchesterheng Posts: 17
in LFW212 Class Forum

Refer to express code to get attacker IP, req.socket.remoteAddress === '127.0.0.1'

The code cannot work as the actual value of req.socket.remoteAddress I get is :ffff:127.0.0.1.

I can process it to remove :ffff:. Anyone encounter this issue?

Comments

  • canelachocanelacho Posts: 12

    Just for local test, use the result when you request "req.socket.remoteAddress", use this value as a string and that's it.

    in your case you are getting the ipv6 and the ipv4 "ffff:127.0.0.1" , in real world you could receive an array of ips in other ways. but this is just an exercise to evaluate.

  • chesterhengchesterheng Posts: 17

    Yes. I process "ffff:127.0.0.1" to remove ":ffff:".

    So you mean I get ":ffff:" because of localhost?
    In real production env, I will get a array of ip without ":ffff:" ?

  • davidmarkclementsdavidmarkclements Posts: 122

    yeah exactly - if your system is setup slightly different you may get an ipv4 inside an ipv6 address. It doesn't matter, the principle is the same.

Sign In or Register to comment.