这里有很多很棒的观点，但没有一个是全面的，所以这里是我最终使用的:

function getIP(req) {
  // req.connection is deprecated
  const conRemoteAddress = req.connection?.remoteAddress
  // req.socket is said to replace req.connection
  const sockRemoteAddress = req.socket?.remoteAddress
  // some platforms use x-real-ip
  const xRealIP = req.headers['x-real-ip']
  // most proxies use x-forwarded-for
  const xForwardedForIP = (() => {
    const xForwardedFor = req.headers['x-forwarded-for']
    if (xForwardedFor) {
      // The x-forwarded-for header can contain a comma-separated list of
      // IP's. Further, some are comma separated with spaces, so whitespace is trimmed.
      const ips = xForwardedFor.split(',').map(ip => ip.trim())
      return ips[0]
    }
  })()
  // prefer x-forwarded-for and fallback to the others
  return xForwardedForIP || xRealIP || sockRemoteAddress || conRemoteAddress
}

也有同样的问题…im也是新的javascript，但我解决了这个与req.connection.remoteAddress;这给了我IP地址(但在ipv6格式::ffff.192.168.0.101)，然后.slice删除前7位数字。

var ip = req.connection.remoteAddress;

if (ip.length < 15) 
{   
   ip = ip;
}
else
{
   var nyIP = ip.slice(7);
   ip = nyIP;
}

以下函数涵盖了所有的情况，将会有所帮助

var ip;
if (req.headers['x-forwarded-for']) {
    ip = req.headers['x-forwarded-for'].split(",")[0];
} else if (req.connection && req.connection.remoteAddress) {
    ip = req.connection.remoteAddress;
} else {
    ip = req.ip;
}console.log("client IP is *********************" + ip);

我在nginx后面使用express和

req.headers.origin

对我有用吗

获取ip地址有两种方式:

让IP = req.ip 让ip = req.connection.remoteAddress;

但上述方法存在一个问题。

如果你在Nginx或任何代理程序后面运行你的应用程序，每个IP地址将是127.0.0.1。

因此，获取user的ip地址的最佳方案是:-

let ip = req.header('x-forwarded-for') || req.connection.remoteAddress;

在你的请求对象中有一个属性叫socket，它是一个网络。套接字对象。净。套接字对象有一个属性remoteAddress，因此你应该能够通过这个调用得到IP:

request.socket.remoteAddress

(如果您的节点版本低于13，请使用已弃用的request.connection.remoteAddress)

EDIT

正如@juand在评论中指出的那样，如果服务器位于代理之后，获得远程IP的正确方法是request.headers['x-forwarded-for']

编辑2

在Node.js中使用express时:

如果你设置了app.set('信任代理'，true)，请请求。ip将返回真实ip地址，即使在代理。查看文档了解更多信息