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

我使用这个ipv4格式

req.connection.remoteAddress.split(':').slice(-1)[0]

其他回答

在shell中,只需curl https://api.ipify.org

所以,让我们观察一下,把它移植到node.js!

Curl是一个从网站获取数据的应用程序,我们传递网站“https://api.ipify.org”作为参数。我们可以使用节点取回来替换curl!

我们从网站上获得的数据是我们的IP地址,这是一种获取你的IP地址的东西。

总结一下:

const fetch = require('node-fetch');

fetch('https://api.ipify.org')
  .then(response => {/* whatever */})
  .catch(err => {/* whatever */})

在nodejs中简单获取远程ip:

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

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; }

Var ipaddress = (req。标题(“x-forwarded-for”)| | req.connection.remoteAddress | | req.socket.remoteAddress | | req.connection.socket.remoteAddress) .split (", ") [0];

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);