CODEWITHSUNDEEP
javascripthtmlregexip-address
jaipaldom
jaipaldom

Reputation: 33

IP addresses with octet wildcards and ranges

I used this pattern to check whether a field's form is an IP address:

function verifyIP (IPvalue) {
    errorString = "";
    theName = "IPaddress";

    var ipPattern = /^(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3})$/;
    var ipArray = IPvalue.match(ipPattern);

    if (IPvalue == "0.0.0.0") {
        errorString = errorString + theName + ': '+IPvalue+' is a special IP address and cannot be used here.';
    } else if (IPvalue == "255.255.255.255") {
        errorString = errorString + theName + ': '+IPvalue+' is a special IP address and cannot be used here.';
    } if (ipArray == null) {
        errorString = errorString + theName + ': '+IPvalue+' is not a valid IP address.';
    } else {
        for (i = 0; i < 4; i++) {
            thisSegment = ipArray[i];
            if (thisSegment > 255) {
                errorString = errorString + theName + ': '+IPvalue+' is not a valid IP address.';
                i = 4;
            }

            if ((i == 0) && (thisSegment > 255)) {
                errorString = errorString + theName + ': '+IPvalue+' is a special IP address and cannot be used here.';
                i = 4;
            }

            if (thisSegment.toString() == "*")
                errorString = "";
            }
        }

        extensionLength = 3;
        if (errorString == "")
            alert ("That is a valid IP address.");
        else
            alert (errorString);
    }
}

But I need to account for the value of the field having an octet(s) with an asterisk '*' or a range '0-255'.

Like for example:

192.168.1.1 --> It will be OK
192.168.*.* --> It will be OK
192.168.2-3.0-128 --> It will be OK
192.168.2-3.* --> It will be OK

Any ideas? Thank you so much!

Upvotes: 3

Views: 1053

Answers (2)

J0e3gan
J0e3gan

Reputation: 8938

For the specific input strings you provided, start with the following:

^(\d{1,3})\.(\d{1,3})\.(\*|(?:\d{1,3}(?:-\d{1,3})?))\.(\*|(?:\d{1,3}(?:-\d{1,3})?))$

Regular expression visualization

Debuggex Demo

In your JavaScript, this would become:

var ipPattern = /^(\d{1,3})\.(\d{1,3})\.(\*|(?:\d{1,3}(?:-\d{1,3})?))\.(\*|(?:\d{1,3}(?:-\d{1,3})?))$/;

You can further eliminate repetition in the pattern of course, but that will make the evolution from what you provided to start even less clear: start with a more verbose, repetitive pattern; get solid positive and negative tests in place; and then refactor to eliminate repetition as needed/desired.

Upvotes: 3

Ben Cook
Ben Cook

Reputation: 1684

There are at least two issues here:

  1. Your regex should match asterisks in the allowed positions. See regex from J0e3gan's answer.

  2. You are looping through indexes 0-3 of the array returned by match. match returns the full string in index 0 and each segment in the following indexes. Therefore, your for loop should be as follows: for (i = 1; i <= 4; i++) { // processing }.

Upvotes: 0

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