CODEWITHSUNDEEP
javaregex
Codey
Codey

Reputation: 1233

Regular expression to check IP

I want to check if an IP address is between 172.16.0.0 and 172.31.255.255

What I tried is this:

Pattern address = Pattern.compile("172.[16-31].[0-255].[0-255]");

But it doesn't work, the compiler throws an error:

Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal character range near index 8
172.[16-31].[0-255].[0-255]
        ^

Since it's an exercise it has to be done with regular expressions.

Upvotes: 0

Views: 125

Answers (2)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

The reason your regex does not work is that the character group [16-31] means

"character 1, any character between 6 and 3, or character 1"

This is definitely not what you wanted to describe. Treating numbers in regex language is difficult - for instance, 16 through 31 is (1[6-9]|2\d|3[01]), i.e. "1 followed by 6 through 9, 2 followed by any digit, or 3 followed by 0 or 1". You will need a similar expression to describe numbers in the 0..255 range: (25[0-5]|2[0-4]\d|[01]?\d\d?).

A lot better approach would be to use InetAddress, which has a getByName method for parsing the address, and lets you examine the bytes of the address with getAddress() method:

byte[] raw = InetAddress.getByName(ipAddrString).getAddress();
boolean valid = raw[0]==172 && raw[1] >= 16 && raw[1] <= 31;

Upvotes: 3

Tim Biegeleisen
Tim Biegeleisen

Reputation: 520878

One option here would be to split the IP address on period and then check to make sure each component is within the ranges you want:

public boolean isIpValid(String input) {
    String[] parts = input.split("\\.");
    int c1 = Integer.parseInt(parts[0]);
    int c2 = Integer.parseInt(parts[1]);
    int c3 = Integer.parseInt(parts[2]);
    int c4 = Integer.parseInt(parts[3]);

    if (c1 == 172 &&
        c2 >= 16 && c2 <= 31 &&
        c3 >= 0 &&  c3 <= 255 &&
        c4 >= 0 &&  c4 <= 255) {
        System.out.println("IP address is valid.");
        return true;
    } else {
        System.out.println("IP address is not valid.");
        return false;
     }
}

Upvotes: 3

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