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

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

对于Underscore.js和Lodash,正确的一行代码是:

var ip = require('underscore')
    .chain(require('os').networkInterfaces())
    .values()
    .flatten()
    .find({family: 'IPv4', internal: false})
    .value()
    .address;

我所知道的就是我想要以192.168开头的IP地址。这段代码会给你:

function getLocalIp() {
    const os = require('os');

    for(let addresses of Object.values(os.networkInterfaces())) {
        for(let add of addresses) {
            if(add.address.startsWith('192.168.')) {
                return add.address;
            }
        }
    }
}

当然,如果你想要一个不同的数字,你可以改变数字。

下面是我获取本地IP地址的实用方法,假设您正在寻找一个IPv4地址,而机器只有一个真实的网络接口。可以很容易地对其进行重构,以返回多接口机器的IP地址数组。

function getIPAddress() {
  var interfaces = require('os').networkInterfaces();
  for (var devName in interfaces) {
    var iface = interfaces[devName];

    for (var i = 0; i < iface.length; i++) {
      var alias = iface[i];
      if (alias.family === 'IPv4' && alias.address !== '127.0.0.1' && !alias.internal)
        return alias.address;
    }
  }
  return '0.0.0.0';
}

运行程序来解析结果似乎有点可疑。这是我用的。

require('dns').lookup(require('os').hostname(), function (err, add, fam) {
  console.log('addr: ' + add);
})

这将返回您的第一个网络接口本地IP地址。