我不完全明白我应该如何获得一个远程用户IP地址。
假设我有一个简单的请求路由，如:
app.get(/, function (req, res){
var forwardedIpsStr = req.header('x-forwarded-for');
var IP = '';
if (forwardedIpsStr) {
IP = forwardedIps = forwardedIpsStr.split(',')[0];
}
});
上面的方法是否正确，以获得真实的用户IP地址或有更好的方法? 那么代理呢?
当前回答
与could-flare, nginx和x-real-ip支持
var user_ip;
if(req.headers['cf-connecting-ip'] && req.headers['cf-connecting-ip'].split(', ').length) {
let first = req.headers['cf-connecting-ip'].split(', ');
user_ip = first[0];
} else {
let user_ip = req.headers['x-forwarded-for'] || req.headers['x-real-ip'] || req.connection.remoteAddress || req.socket.remoteAddress || req.connection.socket.remoteAddress;
}
其他回答
在nginx.conf文件中: proxy_set_header X-Real-IP $remote_addr;
在node.js服务器文件中: Var IP = req。headers['x-real-ip'] || req.connection.remoteAddress;
注意，表示小写头
添加app.set('信任代理'，true) 使用要求。IP或req。Ips和往常一样
如果你在运行一个像NGiNX之类的代理，那么你应该检查'x-forward -for':
var ip = req.headers['x-forwarded-for'] || req.socket.remoteAddress
如果代理不是“你的”，我不会相信“x-forward -for”报头，因为它可能被欺骗。
我为此写了一个包。您可以将其用作表示中间件。我的软件包发布在这里:https://www.npmjs.com/package/express-ip
您可以使用
npm i express-ip
使用
const express = require('express');
const app = express();
const expressip = require('express-ip');
app.use(expressip().getIpInfoMiddleware);
app.get('/', function (req, res) {
console.log(req.ipInfo);
});
与could-flare, nginx和x-real-ip支持
var user_ip;
if(req.headers['cf-connecting-ip'] && req.headers['cf-connecting-ip'].split(', ').length) {
let first = req.headers['cf-connecting-ip'].split(', ');
user_ip = first[0];
} else {
let user_ip = req.headers['x-forwarded-for'] || req.headers['x-real-ip'] || req.connection.remoteAddress || req.socket.remoteAddress || req.connection.socket.remoteAddress;
}
