如何获取Node.js中的本地IP地址？

Question

我不是指

127.0.0.1

反倒是其他计算机将使用例如访问本机的一个

192.168.1.6

Answer 1

http://nodejs.org/api/os.html#os_os_networkinterfaces

var os = require('os'); 

var interfaces = os.networkInterfaces(); 
var addresses = []; 
for (var k in interfaces) { 
    for (var k2 in interfaces[k]) { 
     var address = interfaces[k][k2]; 
     if (address.family === 'IPv4' && !address.internal) { 
      addresses.push(address.address); 
     } 
    } 
} 

console.log(addresses); 

Answer 2

https://github.com/indutny/node-ip

var ip = require("ip"); 
console.dir (ip.address()); 

Answer 3

https://github.com/dominictarr/my-local-ip

$ npm install -g my-local-ip 
$ my-local-ip 

或

$ npm install --save my-local-ip 
$ node 
> console.log(require('my-local-ip')()) 

这是一个很小的模块。

Answer 4

$ npm install --save quick-local-ip 

通过

follwed

var myip = require('quick-local-ip'); 

//getting ip4 network address of local system 
myip.getLocalIP4(); 

//getting ip6 network address of local system 
myip.getLocalIP6(); 

Answer 5

我的版本，这需要一个紧凑和单个文件的脚本，希望能对他人有用：

var ifs = require('os').networkInterfaces(); 
var result = Object.keys(ifs) 
    .map(x => [x, ifs[x].filter(x => x.family === 'IPv4')[0]]) 
    .filter(x => x[1]) 
    .map(x => x[1].address); 

或回答原来的问题：

var ifs = require('os').networkInterfaces(); 
var result = Object.keys(ifs) 
    .map(x => ifs[x].filter(x => x.family === 'IPv4' && !x.internal)[0]) 
    .filter(x => x)[0].address;