IP地址的正则表达式，最后一部分非零

Question

This is my regex for IP address. How can I modify it to exclude a 0 (ZERO) in the last part.

^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.([01]?\\\\d\\\\d?|2[0-4]\\\\d|25[0-5])\\\\.([01]?\\\\d\\\\d?|2[0-4]\\\\d|25[0-5])\\\\.([01]?\\d\\d?|2[0-4]\\d|25[0-5])$

Example:

148.23.54.20 //valid

148.23.54.0 //Invalid

Answer 1

Make sure your number does not start with a zero...

[1-9]\d?

So, for example, your regex would end with

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

This will allow 1 to 255.

The extra escapes were left out for clarity.

Sidenote

For your other 3 cases that that can be 0, simply add an option for only 0

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

Answer 2

仅特殊处理最后一部分

^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?)$

Answer 3

How about a non regex solution:

String ip = /* some ip */;
String[] bytes = ip.split("\\.");
if (!bytes[3].equals("0") && Arrays.stream(bytes).map(Integer::parseInt).allMatch(i -> i >= 0 && i < 256)) {
    //Valid ip
}

You can check for IntegerParseException and do some bounds checks if you're expecting some crazy input, as well. I believe this is a lot more readable and thus maintainable for that sake