如何在不利用其他 IP 地址的情况下从 Ruby 获取我机器的 IP 地址？
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How to get my machine's IP address from Ruby without leveraging from other IP address?
提问by user1535147
I have searched everywhere but their solution requires some form of IP address. Here are the solutions i have found.
我到处搜索，但他们的解决方案需要某种形式的 IP 地址。以下是我找到的解决方案。
require 'socket'
#METHOD 1
ip = IPSocket.getaddress(Socket.gethostname)
puts ip
#METHOD 2
host = Socket.gethostname
puts host
#METHOD 3(uses Google's address)
ip = UDPSocket.open {|s| s.connect("64.233.187.99", 1); s.addr.last}
puts ip
#METHOD 4(uses gateway address)
def local_ip
orig, Socket.do_not_reverse_lookup = Socket.do_not_reverse_lookup, true # turn off reverse DNS resolution temporarily
UDPSocket.open do |s|
s.connect '192.168.1.1', 1
s.addr.last
end
ensure
Socket.do_not_reverse_lookup = orig
end
ip=local_ip
puts ip
All of them require IP address of someone. Is there a solution that does not use someone else's IP address? Preferably, platform independent.
所有这些都需要某人的 IP 地址。有没有不使用别人IP地址的解决方案？最好是平台独立的。
回答by Itay Grudev
Isn't the solution you are looking for just:
您正在寻找的解决方案不只是：
require 'socket'
addr_infos = Socket.ip_address_list
Since a machine can have multiple interfaces and multiple IP Addresses, this method returns an array of Addrinfo.
由于一台机器可以有多个接口和多个 IP 地址，因此此方法返回一个Addrinfo数组。
You can fetch the exact IP addresses like this:
您可以像这样获取确切的 IP 地址：
addr_infos.each do |addr_info|
puts addr_info.ip_address
end
P.S. The question is a bit old, but shows up as the first item on Google and it doesn't contain the solution most people are looking for, so I decided to post it.
PS 这个问题有点老了，但在谷歌上显示为第一项，它不包含大多数人正在寻找的解决方案，所以我决定发布它。
Hope it helps.
希望能帮助到你。
回答by fuzzygroup
This is what I've been using in production for years:
这是我多年来在生产中一直使用的：
require 'socket'
ip = Socket.ip_address_list.detect{|intf| intf.ipv4_private?}
ip.ip_address
Works great; tested on aws and classical hosting
效果很好；在 aws 和经典托管上测试
回答by Eugene Rourke
require 'socket'
Socket::getaddrinfo(Socket.gethostname,"echo",Socket::AF_INET)[0][3]
quite like method 1, actually
很像方法1，实际上
回答by alk
As there is no such thing as a default ip-interface to a host (there does not need to be any ip-interface at all actually) all assumptions regarding nameing are vague, do not necessarily hold.
由于没有主机的默认 ip 接口（实际上根本不需要任何 ip 接口），所有关于命名的假设都是模糊的，不一定成立。
The value returned by gethostname()can be defined independently to any ip-setup, so it does not need to reflect a valid host in terms of a hostname which could be resolved to any ip-address.
返回的值gethostname()可以独立于任何 ip-setup 定义，因此它不需要根据可以解析为任何 ip-address 的主机名来反映有效主机。
From the POSIX system's API's view the only reliabe function to test for the availablily of (ip-)interfaces is the function getifaddrs(), which returns a list of all interfaces along with their parameters.
从 POSIX 系统的 API 的角度来看，唯一用于测试 (ip-) 接口可用性的可靠函数是 function getifaddrs()，它返回所有接口及其参数的列表。
As it looks as if Ruby's current Socket lib does not provide an interface to it, this (http://rubygems.org/gems/system-getifaddrs)gem based approach does seem to be the only way to go.
看起来 Ruby 的当前 Socket 库似乎没有为其提供接口，因此这种 (http://rubygems.org/gems/system-getifaddrs)基于 gem 的方法似乎是唯一的出路。
回答by rogerdpack
parse the output of the ipcommand?
解析ip命令的输出？
from https://gist.github.com/henriquemenezes/a99f13da957515023e78aea30d6c0a48
来自https://gist.github.com/henriquemenezes/a99f13da957515023e78aea30d6c0a48
gw = `ip route show`[/default.*/][/\d+\.\d+\.\d+\.\d+/]
or parse the output of the ipconfigcommand: https://stackoverflow.com/a/12632929/32453
或解析ipconfig命令的输出：https: //stackoverflow.com/a/12632929/32453
