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10 - Block attacker IP with express

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

  • 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.

  • 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:" ?

  • 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.

  • same issue here and i fixed like this:

    app.use(function (req, res, next) {
      const ipList = req.socket.remoteAddress.split(':')
      if (ipList.includes('127.0.0.1')) {
        const err = new Error('Forbidden');
        err.status = 403;
        next(err);
        return;
      }
      next();
    });
    
  • xdxmxc
    xdxmxc Posts: 110

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