用于检查IP的正则表达式

Question

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.

Answer 1

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;
     }
}

Answer 2

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]) , ie " 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;