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1

I'm trying to exclude some internal IP addresses and some internal IP address formats from viewing certain logos and links in the site.I have multiple range of IP addresses(sample given below). Is it possible to write a regex that could match all the IP addresses in the list below using javascript?

10.X.X.X
12.122.X.X
12.211.X.X
64.X.X.X
64.23.X.X
74.23.211.92
and 10 more
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4  
Yes, it is possible: .* matches all these... Your question is vague - do you want any kind of IPv4 address to match? If not, what are the rules for matching? Is that a real X or did you mean this as a placeholder for "anything"? What are the "10 more" IP addresses? – Tim Pietzcker Apr 19 '12 at 21:04
    
I have 15 IP address format(Note: "format" some are exact IP addresses and some that need to match the formats like 10.X.X.X) I need to match. I have provided few samples. – web_dev Apr 19 '12 at 21:10

5 Answers 5

up vote 2 down vote accepted

Quote the periods, replace the X's with [^.]+, and join them all together with pipes:

var allowedIPs = [ 
  "10.X.X.X", 
  "12.122.X.X",
  "12.211.X.X",
  "64.X.X.X",
  "64.23.X.X",
  "74.23.211.92"   //, etc.
]

var allowedPattern = '^(?:' + 
  allowedIPs.join('|').replace(/\./g, '\\.').replace(/X/g, '[^.]+') + 
')$'

 var allowedRegexp = new RegExp(allowedPattern)

Then you're all set:

 '10.1.2.3'.match(allowedRegexp) // => ['10.1.2.3']
 '100.1.2.3'.match(allowedRegexp) // => null

If your input might not even be a legal IP address at all, you can change the '[^.]+' to something that only matches a valid byte value, like '(?:\\d{1,2}|1\\d{2}|2[0-4]\\d|25[0-5])'.

share|improve this answer
    
could you explain what this line of code does? '^(?:' + allowedIPs.join('|').replace(/\./g, '\\.').replace(/X/g, '[^.]+') + ')$' I see it working as per my requirement, very curious how it does.. – web_dev Apr 20 '12 at 14:55
    
It turns the array of allowed IPs into a regular expression that will match any of them. .join('|') combines the array into one big string separated by '|' (e.g. "10.X.X.X|12.122.X.X|..."). The '|' means "or" in a regexp. .replace(/\./g, '\\.') quotes all the periods with backslashes; periods in a regexp match any character, but we want to match only literal periods. .replace(/X/g, '[^.]+') replaces all the 'X's with '[^.]+', which in a regexp means "1 or more non-periods". Putting it all between ^(...)$ means the whole string must match; otherwise "110.2.3.4" would match "10.x.x.x". – Mark Reed Apr 20 '12 at 15:21
up vote 1 down vote

You could do it in regex, but it's not going to be pretty, especially since JavaScript doesn't even support verbose regexes, which means that it has to be one humongous line of regex without any comments. Furthermore, regexes are ill-suited for matching ranges of numbers. I suspect that there are better tools for dealing with this.

Well, OK, here goes (for the samples you provided):

var myregexp = /\b(?:74\.23\.211\.92|(?:12\.(?:122|211)|64\.23)\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])|(?:10|64)\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9]))\b/g;

As a verbose ("readable") regex:

\b                 # start of number
(?:                # Either match...
 74\.23\.211\.92   # an explicit address
|                  # or
 (?:               # an address that starts with
  12\.(?:122|211)  # 12.122 or 12.211
 |                 # or
  64\.23           # 64.23
 )
 \.                # . 
 (?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.  # followed by 0..255 and a dot
 (?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])    # followed by 0..255
|                  # or
 (?:10|64)         # match 10 or 64
 \.                # . 
 (?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.  # followed by 0..255 and a dot
 (?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.  # followed by 0..255 and a dot
 (?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])    # followed by 0..255
)
\b                 # end of number
share|improve this answer
up vote 0 down vote

/^(X|\d{1,3})(\.(X|\d{1,3})){3}$/ should do it.

share|improve this answer
    
could you care to explain what this regex does? – web_dev Apr 19 '12 at 21:11
    
It matches a string in the format X.X.X.X, where X can be 1-3 numeric digits or a capital X. The meat of it is this sub-expression \.(X|\d{1,3}). That matches a period, followed by either 1-3 numeric digits or an X. – cebarrett Apr 19 '12 at 21:20
    
This is a pretty good reference for regex syntax: regular-expressions.info/reference.html – cebarrett Apr 19 '12 at 21:22
    
thanks for the reference. – web_dev Apr 20 '12 at 14:44
up vote 0 down vote

If you don't actually need to match the "X" character you could use this:

\b(?:\d{1,3}\.){3}\d{1,3}\b

Otherwise I would use the solution cebarrett provided.

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up vote 0 down vote

I'm not entirely sure of what you're trying to achieve here (doesn't look anyone else is either).

However, if it's validation, then here's a solution to validate an IP address that doesn't use RegEx. First, split the input string at the dot. Then using parseInt on the number, make sure it isn't higher than 255.

function ipValidator(ipAddress) {
var ipSegments = ipAddress.split('.');
for(var i=0;i<ipSegments.length;i++)
    {
        if(parseInt(ipSegments[i]) > 255){
            return 'fail';
        }
    }
return 'match';
}

Running the following returns 'match':

document.write(ipValidator('10.255.255.125'));

Whereas this will return 'fail':

document.write(ipValidator('10.255.256.125'));

Here's a noted version in a jsfiddle with some examples, http://jsfiddle.net/VGp2p/2/

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