类似于其他答案，但更简洁:

'use strict';

const interfaces = require('os').networkInterfaces();

const addresses = Object.keys(interfaces)
  .reduce((results, name) => results.concat(interfaces[name]), [])
  .filter((iface) => iface.family === 'IPv4' && !iface.internal)
  .map((iface) => iface.address);

你可以通过使用os模块找到你机器的任何IP地址-这是Node.js的本机:

var os = require('os');

var networkInterfaces = os.networkInterfaces();

console.log(networkInterfaces);

你所需要做的就是调用os.networkInterfaces()，你会得到一个容易管理的列表——比按联盟运行ifconfig要简单。

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' } }

下面是前面例子的一个变种。它会小心过滤掉VMware接口等。如果你不传递索引，它会返回所有地址。否则，您可能希望将其默认值设置为0，然后传递null以获取所有值，但您将整理这些。如果想要添加的话，还可以为regex过滤器传入另一个参数。

function getAddress(idx) {

    var addresses = [],
        interfaces = os.networkInterfaces(),
        name, ifaces, iface;

    for (name in interfaces) {
        if(interfaces.hasOwnProperty(name)){
            ifaces = interfaces[name];
            if(!/(loopback|vmware|internal)/gi.test(name)){
                for (var i = 0; i < ifaces.length; i++) {
                    iface = ifaces[i];
                    if (iface.family === 'IPv4' &&  !iface.internal && iface.address !== '127.0.0.1') {
                        addresses.push(iface.address);
                    }
                }
            }
        }
    }

    // If an index is passed only return it.
    if(idx >= 0)
        return addresses[idx];
    return addresses;
}

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

类似于其他答案，但更简洁:

'use strict';

const interfaces = require('os').networkInterfaces();

const addresses = Object.keys(interfaces)
  .reduce((results, name) => results.concat(interfaces[name]), [])
  .filter((iface) => iface.family === 'IPv4' && !iface.internal)
  .map((iface) => iface.address);