对上面答案的改进，原因如下:

Code should be as self-explanatory as possible. Enumerating over an array using for...in... should be avoided. for...in... enumeration should be validated to ensure the object's being enumerated over contains the property you're looking for. As JavaScript is loosely typed and the for...in... can be handed any arbitrary object to handle; it's safer to validate the property we're looking for is available. var os = require('os'), interfaces = os.networkInterfaces(), address, addresses = [], i, l, interfaceId, interfaceArray; for (interfaceId in interfaces) { if (interfaces.hasOwnProperty(interfaceId)) { interfaceArray = interfaces[interfaceId]; l = interfaceArray.length; for (i = 0; i < l; i += 1) { address = interfaceArray[i]; if (address.family === 'IPv4' && !address.internal) { addresses.push(address.address); } } } } console.log(addresses);

调用ifconfig是非常依赖于平台的，而且网络层知道套接字所在的IP地址，所以最好是询问它。

Node.js并没有公开这样做的直接方法，但是你可以打开任何套接字，并询问正在使用的本地IP地址。例如，打开www.google.com的套接字:

var net = require('net');
function getNetworkIP(callback) {
  var socket = net.createConnection(80, 'www.google.com');
  socket.on('connect', function() {
    callback(undefined, socket.address().address);
    socket.end();
  });
  socket.on('error', function(e) {
    callback(e, 'error');
  });
}

使用情况:

getNetworkIP(function (error, ip) {
    console.log(ip);
    if (error) {
        console.log('error:', error);
    }
});

下面是一个允许你获取本地IP地址的变体(在Mac和Windows上测试):

var
    // Local IP address that we're trying to calculate
    address
    // Provides a few basic operating-system related utility functions (built-in)
    ,os = require('os')
    // Network interfaces
    ,ifaces = os.networkInterfaces();


// Iterate over interfaces ...
for (var dev in ifaces) {

    // ... and find the one that matches the criteria
    var iface = ifaces[dev].filter(function(details) {
        return details.family === 'IPv4' && details.internal === false;
    });

    if(iface.length > 0)
        address = iface[0].address;
}

// Print the result
console.log(address); // 10.25.10.147

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';
}

对于任何对简洁感兴趣的人来说，这里有一些“一行程序”，它们不需要不是标准Node.js安装的一部分的插件/依赖项:

eth0的公共IPv4、IPv6地址为阵列:

var ips = require('os').networkInterfaces().eth0.map(function(interface) {
    return interface.address;
});

eth0的第一个公网IP地址(一般为IPv4):

var ip = require('os').networkInterfaces().eth0[0].address;