Private IP Address Identifier in Regular Expressio

问题:

I'm wondering if this is the best way to match a string that starts with a private IP address (Perl-style Regex):

(^127\.0\.0\.1)|(^192\.168)|(^10\.)|(^172\.1[6-9])|(^172\.2[0-9])|(^172\.3[0-1])

Thanks much!

回答1:

I'm assuming you want to match these ranges:

127.  0.0.0 – 127.255.255.255     127.0.0.0 /8
 10.  0.0.0 –  10.255.255.255      10.0.0.0 /8
172. 16.0.0 – 172. 31.255.255    172.16.0.0 /12
192.168.0.0 – 192.168.255.255   192.168.0.0 /16

You are missing some dots that would cause it to accept for example 172.169.0.0 even though this should not be accepted. I've fixed it below. Remove the new lines, it's just for readability.

(^127\.)|
(^10\.)|
(^172\.1[6-9]\.)|(^172\.2[0-9]\.)|(^172\.3[0-1]\.)|
(^192\.168\.)

Also note that this assumes that the IP addresses have already been validated - it accepts things like 10.foobar.

回答2:

This is the same as the correct answer by Mark, but now including IPv6 private addresses.

/(^127\.)|(^192\.168\.)|(^10\.)|(^172\.1[6-9]\.)|(^172\.2[0-9]\.)|(^172\.3[0-1]\.)|(^::1$)|(^[fF][cCdD])/

回答3:

I have generated this

REGEXP FOR CLASS A NETWORKS :

(10)(\.([2]([0-5][0-5]|[01234][6-9])|[1][0-9][0-9]|[1-9][0-9]|[0-9])){3}

REGEXP FOR CLASS B NETWORKS :

(172)\.(1[6-9]|2[0-9]|3[0-1])(\.([2][0-5][0-5]|[1][0-9][0-9]|[1-9][0-9]|[0-9])){2}

REGEXP FOR CLASS C NETWORKS :

(192)\.(168)(\.[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5]){2}

Let me know if you encounter any error

If you are sure of your output (say for example netstat) and you have no need to check about IP address validity because it is already done, then you can catch private ip addresses with this formula

grep -P "(10.|192.168|172.1[6-9].|172.2[0-9].|172.3[01].).* "

回答4:

This is in case you decide to go with my comment, suggesting you don't use regexps. Untested (but probably works, or at least close), in Perl:

@private = (
    {network => inet_aton('127.0.0.0'),   mask => inet_aton('255.0.0.0')   },
    {network => inet_aton('192.168.0.0'), mask => inet_aton('255.255.0.0') },
    # ...
);

$ip = inet_aton($ip_text);
if (grep $ip & $_->{mask} == $_->{network}, @private) {
    # ip address is private
} else {
    # ip address is not private
}

Note now how @private is just data, which you can easily change. Or download on the fly from the Cymru Bogon Reference.

edit: It occurs to me that asking for a Perl regexp doesn't mean you know Perl, so the key line is there is the 'grep', which just loops over each private address range. You take your IP, bitwise and it with the netmask, and compare to the network address. If equal, its part of that private network.

回答5:

Looks right. Personally, I'd change the first one to:

^127\.0 

With this: (^127\.0\.0\.1) you looking for anything that starts with 127.0.0.1 and will miss out on 127.0.0.2*, 127.0.2.*, 127.0.* etc.

回答6:

If you're looking for system.net defaultProxy and proxy bypasslist config that uses a proxy for external but uses direct connections for internal hosts (could do with some ipv6 support)...

<system.net>
  <defaultProxy enabled="true">
    <proxy proxyaddress="http://proxycluster.privatedomain.net:8080" bypassonlocal="True"  />
    <bypasslist>
      <!-- exclude local host -->
      <add address="^(http|https)://localhost$" />
      <!-- excludes *.privatedomain.net -->
      <add address="^(http|https)://.*\.privatedomain\.net$" />
      <!-- excludes simple host names -->
      <add address="^(http|https)://[a-z][a-z0-9\-_]*$" />
      <!-- exclude private network addresses 192.168, 172.16..31 through 31, 127.* etc. -->
      <add address="^(http|https)://((((127)|(10))\.[0-9]+\.[0-9]+\.[0-9]+)|(((172\.(1[6-9]|2[0-9]|3[0-1]))|(192\.168))\.[0-9]+\.[0-9]+))$"/>
    </bypasslist>
  </defaultProxy>
  <connectionManagement>
    <add address="*" maxconnection="10" />
  </connectionManagement>
</system.net>

回答7:

here is what I use in python:

rfc1918 = re.compile('^(10(\.(25[0-5]|2[0-4][0-9]|1[0-9]{1,2}|[0-9]{1,2})){3}|((172\.(1[6-9]|2[0-9]|3[01]))|192\.168)(\.(25[0-5]|2[0-4][0-9]|1[0-9]{1,2}|[0-9]{1,2})){2})$')

You can remove the ^ and/or $ anchors if you wish.

I prefer the above regex because it weeds out invalid octets (anything above 255).

example usage:

if rfc1918.match(ip):
    print "ip is private"

回答8:

     //RegEx to check for the following ranges. IPv4 only
         //172.16-31.xxx.xxx
         //10.xxx.xxx.xxx
         //169.254.xxx.xxx
         //192.168.xxx.xxx

     var regex = /(^127\.)|(^(0)?10\.)|(^172\.(0)?1[6-9]\.)|(^172\.(0)?2[0-9]\.)|(^172\.(0)?3[0-1]\.)|(^169\.254\.)|(^192\.168\.)/;

回答9:

FWIW this pattern was over 10% faster using pattern.matcher:

^1((0)|(92\\.168)|(72\\.((1[6-9])|(2[0-9])|(3[0-1])))|(27))\\.