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
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);
其他回答
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;
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
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());
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';
}
推荐文章
- 我如何使用Jest模拟JavaScript的“窗口”对象?
- 我如何等待一个承诺完成之前返回一个函数的变量?
- CALL_AND_RETRY_LAST分配失败-进程内存不足
- 在JavaScript中根据键值查找和删除数组中的对象
- 使嵌套JavaScript对象平放/不平放的最快方法
- 在Ubuntu上安装Node.js
- 如何以及为什么'a'['toUpperCase']()在JavaScript工作?
- 有Grunt生成index.html不同的设置
- 文档之间的区别。addEventListener和window。addEventListener?
- 如何检查动态附加的事件监听器是否存在?
- 使用express.js代理
- 如何写setTimeout与参数Coffeescript
- 将JavaScript字符串中的多个空格替换为单个空格
- JavaScript: override alert()
- 重置setTimeout