如何从控制器内确定给定请求的IP地址?例如(在快递中):

app.post('/get/ip/address', function (req, res) {
    // need access to IP address here
})

当前回答

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;
}

其他回答

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);

函数getCallerIP(请求){ Var IP = request。标题(“x-forwarded-for”)| | request.connection.remoteAddress | | request.socket.remoteAddress | | request.connection.socket.remoteAddress; IP = IP .split(',')[0]; IP = IP .split(':').slice(-1);//如果IP以“::ffff:146.xxx.xxx.xxx”格式返回 返回的ip; }

对于我使用kubernetes ingress (NGINX):

req.headers['x-original-forwarded-for']

在Node.js中非常有效

如果使用express…

req.ip

我在查这个,然后我想,等等,我用的是快递。咄。

我们可以在node js中检查这段代码

const os       = require('os');
const interfaces = os.networkInterfaces();

let addresses = [];

for (var k in interfaces) {

    for (var k2 in interfaces[k]) {

        const address = interfaces[k][k2];

        if ( (address.family === 'IPv4' || address.family === 'IPv6')  && 
            !address.internal) {

            addresses.push(address.address);

        }
    }
}
console.log(addresses);