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

如果你喜欢简洁的东西,下面是使用Lodash:

Var OS = require(' OS '); Var _ = require('lodash'); var firstLocalIp = _(os.networkInterfaces()).values().flatten().where({family: 'IPv4', internal: false}).pluck('address').first(); console.log('第一个本地IPv4地址是' + firstLocalIp);

其他回答

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);

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

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';
}

下面的解决方案对我来说是可行的

const ip = Object.values(require("os").networkInterfaces())
        .flat()
        .filter((item) => !item.internal && item.family === "IPv4")
        .find(Boolean).address;

安装一个名为ip的模块,如下:

npm install ip

然后使用下面的代码:

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