Question

这是一个棘手的问题......

如何知道ip，62.156.244.13是否在62.0.0.0和62.255.255.255

的范围内

有没有一种简单的Ruby方法可以做到这一点，或者我该如何解决这个问题呢？

Answer 1

>> require "ipaddr"
=> true
>> low = IPAddr.new("62.0.0.0").to_i
=> 1040187392
>> high = IPAddr.new("62.255.255.255").to_i
=> 1056964607
>> ip = IPAddr.new("62.156.244.13").to_i
=> 1050473485
>> (low..high)===ip
=> true

如果您获得网络而不是起始地址和结束地址，则更简单

>> net = IPAddr.new("62.0.0.0/8")
=> #<IPAddr: IPv4:62.0.0.0/255.0.0.0>
>> net===IPAddr.new("62.156.244.13")
=> true

IPAddr也可以使用IPv6地址

>> low = IPAddr.new('1::')
=> #<IPAddr: IPv6:0001:0000:0000:0000:0000:0000:0000:0000/ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff>
>> high = IPAddr.new('2::')
=> #<IPAddr: IPv6:0002:0000:0000:0000:0000:0000:0000:0000/ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff>
>> (low..high)===IPAddr.new('1::1')
=> true

Answer 2

我会使用this killer little function将IP地址转换为整数，然后进行比较。

def ip_addr_in_range?(low, high, addr)
  int_addr = numeric_ip(addr)
  int_addr <= numeric_ip(high) && int_addr >= numeric_ip(low)
end

def numeric_ip(ip_str)
  ip_str.split('.').inject(0) { |ip_num, part| ( ip_num << 8 ) + part.to_i }
end

def test_ip_addr_in_range(low, high, addr, expected)
  result = ip_addr_in_range?(low, high, addr)
  puts "#{addr} #{(expected ? 'should' : 'should not')} be within #{low} and #{high}: #{(expected == result ? 'PASS' : 'FAIL')}"
end


test_ip_addr_in_range("192.168.0.0", "192.168.0.255", "192.168.0.200", true)
test_ip_addr_in_range("192.168.0.0", "192.168.0.155", "192.168.0.200", false)
test_ip_addr_in_range("192.168.0.0", "192.168.255.255", "192.168.100.200", true)
test_ip_addr_in_range("192.168.0.0", "192.168.100.255", "192.168.150.200", false)
test_ip_addr_in_range("192.168.255.255", "192.255.255.255", "192.200.100.100", true)
test_ip_addr_in_range("192.168.255.255", "192.255.255.255", "192.100.100.100", false)
test_ip_addr_in_range("192.168.255.255", "255.255.255.255", "200.200.100.100", true)
test_ip_addr_in_range("192.168.255.255", "255.255.255.255", "180.100.100.100", false)

$ ruby ip_range.rb
192.168.0.200 should be within 192.168.0.0 and 192.168.0.255: PASS
192.168.0.200 should not be within 192.168.0.0 and 192.168.0.155: PASS
192.168.100.200 should be within 192.168.0.0 and 192.168.255.255: PASS
192.168.150.200 should not be within 192.168.0.0 and 192.168.100.255: PASS
192.200.100.100 should be within 192.168.255.255 and 192.255.255.255: PASS
192.100.100.100 should not be within 192.168.255.255 and 192.255.255.255: PASS
200.200.100.100 should be within 192.168.255.255 and 255.255.255.255: PASS
180.100.100.100 should not be within 192.168.255.255 and 255.255.255.255: PASS

Answer 3

受到jdl答案的启发，但有一个范围：

class String
  def to_ip
    split(".").inject(0) { |s, p| (s << 8) + p.to_i }
  end
end

("62.0.0.0".to_ip.."62.255.255.255".to_ip).include?("62.156.244.13".to_ip)

Answer 4

有一种方法#include?

您可以这样做：

IPAddr.new("127.0.0.1/8").include? "127.1.10.200"

Answer 5

我更喜欢这种方法将IP地址转换为整数以进行范围比较：

# n_ip("192.1.1.23") will return the number 192001001023
def n_ip(input)
  input.split(".").collect{|p| p.rjust(3, "0")}.join.to_i
end

def ip_in_range?(from, to, given)
  (n_ip(from)..n_ip(to).include?(n_ip(given))
end

现在您可以按如下方式检查值：

>> ip_in_range?("192.168.0.0",  "192.168.0.255","192.168.0.200")
ip_in_range?("192.168.0.0",  "192.168.0.255","192.168.0.200")
=> true
>> ip_in_range?("192.168.0.0",  "192.168.0.255", "192.168.1.200")
ip_in_range?("192.168.0.0",  "192.168.0.255", "192.168.1.200")
=> false

返回的整数不是IP地址的32位表示。该逻辑适用于所有有效的IP地址。

Answer 6

jdl的answer很好。我会做in_range？功能一线：

def ip_addr_in_range?(low, high, addr)
  (numeric_ip(low) .. numeric_ip(high)) === numeric_ip(addr)
end

Answer 7

不要让它变得比它更难。

def check_ip(ip)
  '62.0.0.0' < ip and ip < '62.255.255.255'
end

check_ip '62.156.244.13' #=> true

编辑：或者，如果您使用的是Rails / ActiveSupport：

def check_ip(ip)
  ip.starts_with? '62'
end

check_ip '62.156.244.13' #=> true

Rails：查明IP是否在IP范围内

7 个答案: