我有一个简单的Node.js程序在我的机器上运行,我想获得我的程序正在运行的PC的本地IP地址。我如何在Node.js中获得它?
当前回答
var ip = req.headers['x-forwarded-for'] || req.socket.remoteAddress
其他回答
这是我的变体,允许以可移植的方式获得IPv4和IPv6地址:
/**
* Collects information about the local IPv4/IPv6 addresses of
* every network interface on the local computer.
* Returns an object with the network interface name as the first-level key and
* "IPv4" or "IPv6" as the second-level key.
* For example you can use getLocalIPs().eth0.IPv6 to get the IPv6 address
* (as string) of eth0
*/
getLocalIPs = function () {
var addrInfo, ifaceDetails, _len;
var localIPInfo = {};
//Get the network interfaces
var networkInterfaces = require('os').networkInterfaces();
//Iterate over the network interfaces
for (var ifaceName in networkInterfaces) {
ifaceDetails = networkInterfaces[ifaceName];
//Iterate over all interface details
for (var _i = 0, _len = ifaceDetails.length; _i < _len; _i++) {
addrInfo = ifaceDetails[_i];
if (addrInfo.family === 'IPv4') {
//Extract the IPv4 address
if (!localIPInfo[ifaceName]) {
localIPInfo[ifaceName] = {};
}
localIPInfo[ifaceName].IPv4 = addrInfo.address;
} else if (addrInfo.family === 'IPv6') {
//Extract the IPv6 address
if (!localIPInfo[ifaceName]) {
localIPInfo[ifaceName] = {};
}
localIPInfo[ifaceName].IPv6 = addrInfo.address;
}
}
}
return localIPInfo;
};
下面是同一个函数的CoffeeScript版本:
getLocalIPs = () =>
###
Collects information about the local IPv4/IPv6 addresses of
every network interface on the local computer.
Returns an object with the network interface name as the first-level key and
"IPv4" or "IPv6" as the second-level key.
For example you can use getLocalIPs().eth0.IPv6 to get the IPv6 address
(as string) of eth0
###
networkInterfaces = require('os').networkInterfaces();
localIPInfo = {}
for ifaceName, ifaceDetails of networkInterfaces
for addrInfo in ifaceDetails
if addrInfo.family=='IPv4'
if !localIPInfo[ifaceName]
localIPInfo[ifaceName] = {}
localIPInfo[ifaceName].IPv4 = addrInfo.address
else if addrInfo.family=='IPv6'
if !localIPInfo[ifaceName]
localIPInfo[ifaceName] = {}
localIPInfo[ifaceName].IPv6 = addrInfo.address
return localIPInfo
console.log(getLocalIPs())的示例输出
{ lo: { IPv4: '127.0.0.1', IPv6: '::1' },
wlan0: { IPv4: '192.168.178.21', IPv6: 'fe80::aa1a:2eee:feba:1c39' },
tap0: { IPv4: '10.1.1.7', IPv6: 'fe80::ddf1:a9a1:1242:bc9b' } }
我可能在这个问题上迟到了,但如果有人想要一个一行ES6解决方案来获得IP地址数组,那么这应该会帮助你:
Object.values(require("os").networkInterfaces())
.flat()
.filter(({ family, internal }) => family === "IPv4" && !internal)
.map(({ address }) => address)
As
Object.values(require("os").networkInterfaces())
将返回一个数组的数组,所以flat()是用来将其平展为单个数组
.filter(({ family, internal }) => family === "IPv4" && !internal)
将过滤数组只包括IPv4地址,如果它不是内部
最后
.map(({ address }) => address)
是否只返回过滤数组的IPv4地址
所以结果是['192.168.xx。xx ']
然后,如果您想要或更改筛选条件,您可以获得该数组的第一个索引
操作系统为Windows
在我看来,这里的一些答案似乎不必要地过于复杂。 这里有一个更好的方法,使用普通的Nodejs。
import os from "os";
const machine = os.networkInterfaces()["Ethernet"].map(item => item.family==="IPv4")
console.log(machine.address) //gives 192.168.x.x or whatever your local address is
参见文档:NodeJS - os模块:networkInterfaces
下面是一段Node.js代码,它将解析ifconfig的输出并(异步地)返回找到的第一个IP地址:
(它只在Mac OS X v10.6 (Snow Leopard)上测试;我希望它也能在Linux上运行。)
var getNetworkIP = (function () {
var ignoreRE = /^(127\.0\.0\.1|::1|fe80(:1)?::1(%.*)?)$/i;
var exec = require('child_process').exec;
var cached;
var command;
var filterRE;
switch (process.platform) {
// TODO: implement for OSes without the ifconfig command
case 'darwin':
command = 'ifconfig';
filterRE = /\binet\s+([^\s]+)/g;
// filterRE = /\binet6\s+([^\s]+)/g; // IPv6
break;
default:
command = 'ifconfig';
filterRE = /\binet\b[^:]+:\s*([^\s]+)/g;
// filterRE = /\binet6[^:]+:\s*([^\s]+)/g; // IPv6
break;
}
return function (callback, bypassCache) {
// Get cached value
if (cached && !bypassCache) {
callback(null, cached);
return;
}
// System call
exec(command, function (error, stdout, sterr) {
var ips = [];
// Extract IP addresses
var matches = stdout.match(filterRE);
// JavaScript doesn't have any lookbehind regular expressions, so we need a trick
for (var i = 0; i < matches.length; i++) {
ips.push(matches[i].replace(filterRE, '$1'));
}
// Filter BS
for (var i = 0, l = ips.length; i < l; i++) {
if (!ignoreRE.test(ips[i])) {
//if (!error) {
cached = ips[i];
//}
callback(error, ips[i]);
return;
}
}
// Nothing found
callback(error, null);
});
};
})();
使用的例子:
getNetworkIP(function (error, ip) {
console.log(ip);
if (error) {
console.log('error:', error);
}
}, false);
如果第二个参数为true,函数将每次执行一次系统调用;否则使用缓存的值。
更新版本
返回所有本地网络地址的数组。
在Ubuntu 11.04 (Natty Narwhal)和Windows XP 32上测试
var getNetworkIPs = (function () {
var ignoreRE = /^(127\.0\.0\.1|::1|fe80(:1)?::1(%.*)?)$/i;
var exec = require('child_process').exec;
var cached;
var command;
var filterRE;
switch (process.platform) {
case 'win32':
//case 'win64': // TODO: test
command = 'ipconfig';
filterRE = /\bIPv[46][^:\r\n]+:\s*([^\s]+)/g;
break;
case 'darwin':
command = 'ifconfig';
filterRE = /\binet\s+([^\s]+)/g;
// filterRE = /\binet6\s+([^\s]+)/g; // IPv6
break;
default:
command = 'ifconfig';
filterRE = /\binet\b[^:]+:\s*([^\s]+)/g;
// filterRE = /\binet6[^:]+:\s*([^\s]+)/g; // IPv6
break;
}
return function (callback, bypassCache) {
if (cached && !bypassCache) {
callback(null, cached);
return;
}
// System call
exec(command, function (error, stdout, sterr) {
cached = [];
var ip;
var matches = stdout.match(filterRE) || [];
//if (!error) {
for (var i = 0; i < matches.length; i++) {
ip = matches[i].replace(filterRE, '$1')
if (!ignoreRE.test(ip)) {
cached.push(ip);
}
}
//}
callback(error, cached);
});
};
})();
使用举例:升级版本
getNetworkIPs(function (error, ip) {
console.log(ip);
if (error) {
console.log('error:', error);
}
}, false);
运行程序来解析结果似乎有点可疑。这是我用的。
require('dns').lookup(require('os').hostname(), function (err, add, fam) {
console.log('addr: ' + add);
})
这将返回您的第一个网络接口本地IP地址。
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