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
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' } }
其他回答
安装一个名为ip的模块,如下:
npm install ip
然后使用下面的代码:
var ip = require("ip");
console.log(ip.address());
下面是一个简单的JavaScript版本,用于获取单个IP地址:
function getServerIp() {
var os = require('os');
var ifaces = os.networkInterfaces();
var values = Object.keys(ifaces).map(function(name) {
return ifaces[name];
});
values = [].concat.apply([], values).filter(function(val){
return val.family == 'IPv4' && val.internal == false;
});
return values.length ? values[0].address : '0.0.0.0';
}
如果你不想安装依赖,并且正在运行*nix系统,你可以这样做:
hostname -I
你会得到主机的所有地址,你可以在node中使用这个字符串:
const exec = require('child_process').exec;
let cmd = "hostname -I";
exec(cmd, function(error, stdout, stderr)
{
console.log(stdout + error + stderr);
});
是一行代码,你不需要像'os'或'node-ip'这样可能会意外增加代码复杂性的其他库。
hostname -h
也是你的朋友;-)
希望能有所帮助!
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
下面的解决方案对我来说是可行的
const ip = Object.values(require("os").networkInterfaces())
.flat()
.filter((item) => !item.internal && item.family === "IPv4")
.find(Boolean).address;