Java Regex mimicking if-else for numbers

Question

Using this as a guide to attempt to emulate an if-else Java regex, I came up with: [0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9])) to do the following:

An optional digit between 0-2 inclusive as the leftmost digit; However, if the first digit is a 2, then the next digit to the right can be maximum 5. If it is a 0 or 1, or left blank, then 0-9 is valid. I am trying to ultimately end up allowing a user to only write the numbers 0-255.

Testing the regular expression on both Regex101 as well as javac doesn't work on test cases, despite the Regex101 explanation being congruent with what I want.

When I test the regex:

System.out.println("0".matches("[0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9]))")); ---> false

System.out.println("2".matches("[0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9]))")); ----> true

System.out.println("25".matches("[0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9]))")); ----> false

System.out.println("22".matches("[0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9]))")); ----> false

System.out.println("1".matches("[0-2]?(?:(?<=2)(?![6-9])|(?<!2)(?=[0-9]))")); ----> false

It appears so far, from few test cases, 2 is the only valid case that is accepted by the regex.

For reference, here is my initial regex, using if-else that limits a number to the range of 0-255: [0-2]?(?(?<=2)[0-5]|[0-9])(?(?<=25)[0-5]|[0-9])

@ScaryWombat There's no code necessary in this question; the regex covers that. And doesn't work fairly obviously means doesn't match expected strings. — FThompson, Feb 10 '16 at 02:08
With that said, though, it'd be nice to include some test cases. — FThompson, Feb 10 '16 at 02:08
@Vulcan If it works with Perl but not Java, then I suggest that some code is necessary. — Scary Wombat, Feb 10 '16 at 02:09
@ScaryWombat In Perl, OP used an if-else. Java doesn't support conditional regexes, so OP is converting the regex to work in Java (unsuccessfully). — FThompson, Feb 10 '16 at 02:11

score 2 · Accepted Answer · edited May 23 '17 at 12:31

2

I don't see why to mimic if else for checking a range. It's just putting some patterns together.

^(?:[1-9]?\d|1\d\d|2[0-4]\d|25[0-5])$

^ start anchor
(?: opens a non capture group for alternation
[1-9]?\d matches 0-99
1\d\d matches 100-199
2[0-4]\d matches 200-249
25[0-5] matches 250-255
$ end anchor

See demo at regex101

With allowing leading zeros, you can reduce it to
^(?:[01]?\d\d?|2[0-4]\d|25[0-5])$

edited May 23 '17 at 12:31

Community

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answered Feb 10 '16 at 03:07

bobble bubble

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It isn't exactly what I wanted, as I wanted cases such as 000 or 001 or 02 to be accepted, but I modified it by prepending: `0\d|00\d|0\d\d|` – Abdul Feb 10 '16 at 04:03

FThompson · Answer 2 · 2016-02-10T02:14:48.913

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As you are trying to only allow a range of numbers (0-255), why use regex at all? Instead, parse the string as an int and check if it falls within the range.

public static boolean isInRange(String input, int min, int max) {
    try {
        int val = Integer.parseInt(input);
        return val >= min && val < max;
    } catch (NumberFormatException e) {
        return false;
    }
}

edited Feb 10 '16 at 02:14

answered Feb 10 '16 at 02:12

FThompson

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Unfortunately the purpose of the exercise is to use Regex only, even if there are much easier and better ways, otherwise it'd be cheating. – Abdul Feb 10 '16 at 02:13

Java Regex mimicking if-else for numbers

2 Answers2