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
})
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
在Typescript中使用ValidatorJS。下面是NodeJS中间件:
// Extract Client IP Address
app.use((req, res, next) => {
let ipAddress = (req.headers['x-forwarded-for'] as string || '').split(',')[0]
if (!validator.isIP(ipAddress))
ipAddress = req.socket.remoteAddress?.toString().split(':').pop() || ''
if (!validator.isIP(ipAddress))
return res.status(400).json({errorMessage: 'Bad Request'})
req.headers['x-forwarded-for'] = ipAddress
next()
})
在这里,我假设所有请求都应该有一个有效的IP地址,因此如果没有找到有效的IP地址,则返回一个代码为400的响应。
其他回答
如果你有多个ip,这对我来说是可行的:
Var ipaddress = (req。标题(“x-forwarded-for”)| | req.connection.remoteAddress | | req.socket.remoteAddress | | req.connection.socket.remoteAddress) .split (", ") [0];
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
Var ipaddress = (req。标题(“x-forwarded-for”)| | req.connection.remoteAddress | | req.socket.remoteAddress | | req.connection.socket.remoteAddress) .split (", ") [0];
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
}
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);