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

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。

其他回答

这些信息可以在os.networkInterfaces()中找到,这是一个对象,它将网络接口名称映射到它的属性(例如,一个接口可以有几个地址):

'use strict';

const { networkInterfaces } = require('os');

const nets = networkInterfaces();
const results = Object.create(null); // Or just '{}', an empty object

for (const name of Object.keys(nets)) {
    for (const net of nets[name]) {
        // Skip over non-IPv4 and internal (i.e. 127.0.0.1) addresses
        // 'IPv4' is in Node <= 17, from 18 it's a number 4 or 6
        const familyV4Value = typeof net.family === 'string' ? 'IPv4' : 4
        if (net.family === familyV4Value && !net.internal) {
            if (!results[name]) {
                results[name] = [];
            }
            results[name].push(net.address);
        }
    }
}
// 'results'
{
  "en0": [
    "192.168.1.101"
  ],
  "eth0": [
    "10.0.0.101"
  ],
  "<network name>": [
    "<ip>",
    "<ip alias>",
    "<ip alias>",
    ...
  ]
}
// results["en0"][0]
"192.168.1.101"

我只用Node.js就能做到这一点。

node . js:

var os = require( 'os' );
var networkInterfaces = Object.values(os.networkInterfaces())
    .reduce((r,a) => {
        r = r.concat(a)
        return r;
    }, [])
    .filter(({family, address}) => {
        return family.toLowerCase().indexOf('v4') >= 0 &&
            address !== '127.0.0.1'
    })
    .map(({address}) => address);
var ipAddresses = networkInterfaces.join(', ')
console.log(ipAddresses);

作为Bash脚本(需要安装Node.js)

function ifconfig2 ()
{
    node -e """
        var os = require( 'os' );
        var networkInterfaces = Object.values(os.networkInterfaces())
            .reduce((r,a)=>{
                r = r.concat(a)
                return r;
            }, [])
            .filter(({family, address}) => {
                return family.toLowerCase().indexOf('v4') >= 0 &&
                    address !== '127.0.0.1'
            })
            .map(({address}) => address);
        var ipAddresses = networkInterfaces.join(', ')
        console.log(ipAddresses);
    """
}

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。

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

很多时候,我发现有多个内部和外部面向接口可用(例如:10.0.75.1,172.100.0.1,192.168.2.3),而我真正想要的是外部接口(172.100.0.1)。

如果其他人也有类似的担忧,这里还有一个关于这个问题的看法,希望能有所帮助……

const address = Object.keys(os.networkInterfaces())
    // flatten interfaces to an array
    .reduce((a, key) => [
        ...a,
        ...os.networkInterfaces()[key]
    ], [])
    // non-internal ipv4 addresses only
    .filter(iface => iface.family === 'IPv4' && !iface.internal)
    // project ipv4 address as a 32-bit number (n)
    .map(iface => ({...iface, n: (d => ((((((+d[0])*256)+(+d[1]))*256)+(+d[2]))*256)+(+d[3]))(iface.address.split('.'))}))
    // set a hi-bit on (n) for reserved addresses so they will sort to the bottom
    .map(iface => iface.address.startsWith('10.') || iface.address.startsWith('192.') ? {...iface, n: Math.pow(2,32) + iface.n} : iface)
    // sort ascending on (n)
    .sort((a, b) => a.n - b.n)
    [0]||{}.address;