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I am facing problem while matching input string with Regex. I want to validate input number is between 0-255 and length should be up to 3 characters long. code is working fine but when I input 000000 up to any length is shows true instead false.
Here is my code :-
String IP = "000000000000000";
System.out.println(IP.matches("(0*(?:[0-9][0-9]?|[0-2][0-5][0-5]))"));
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With regex you have a couple of options to match a digit. You can use a number from 0 to 9 to match a single choice. Or you can match a range of digits with a character group e.g. [4-9]. If the character group allows any digit (i.e. [0-9]), it can be replaced with a shorthand (\d).
If the if part evaluates to true, then the regex engine will attempt to match the then part. Otherwise, the else part is attempted instead. The syntax consists of a pair of parentheses. The opening bracket must be followed by a question mark, immediately followed by the if part, immediately followed by the then part.
Using Regular Expressionregex = "[0-9]+"; Match the given string with Regular Expression. In Java, this can be done by using Pattern. matcher().
You can use regex (. *\d)([A-Z])(\d. *) - This will give you exact ONE Alphabet between numbers.
You can use this pattern which matches "0", "1", ... "255":
"([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])"
Demo on Ideone
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Tested this:
static String pattern = "^(([0-1]?[0-9]?[0-9]?|2[0-4][0-9]|25[0-5])\\.){3}([0-1]?[0-9]?[0-9]?|2[0-4][0-9]|25[0-5]){1}$";
It works for the following:
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