CODEWITHSUNDEEP
java
HARSHIT DANG
HARSHIT DANG

Reputation: 31

Why doesn't Pattern.matches("[a*mn]","aaaa") return true? What should be proper code to get the desired output?

I want to create a pattern where the desired string should either be multiples of a including null i.e. a*, or it should be one single m or single n. But the following code doesn't give the desired output.

class Solution {
            public static void main(String args[]) {
                System.out.println(Pattern.matches("[a*mn]", "aaaa"));
        }
}

Upvotes: 1

Views: 126

Answers (3)

Syed Salman Hassan
Syed Salman Hassan

Reputation: 32

inside the [] the "" is not a quantifier so you'll get a true if one of the characters in the regex is present therefore the result will be true if the string is "a","","m" or "n". And the rest will result in false. your regex should be:

([aa*]*|[mn])

it will be true only if multiples of "a" are entered including "a*" or a single "m" or "n". check it by following examples:

System.out.println("[aa*]*|[mn]","m");
System.out.println("[aa*]*|[mn]","aaaaa");
System.out.println("[aa*]*|[mn]","a*a*");

Upvotes: 1

T.J. Crowder
T.J. Crowder

Reputation: 1074248

* within a character class ([]) is just a *, not a quantifier.

I want to create a pattern where the desired string should either be multiples of a including null i.e. a*, or it should be one single m or single n.

You'll need an alternation (|) for that: a*|[mn]:

Pattern.matches("a*|[mn]", "aaaa")

Live example:

import java.util.regex.Pattern;

class Example {
    public static void main (String[] args) throws java.lang.Exception {
        check("aaaa", true);
        check("a", true);
        check("", true);
        check("m", true);
        check("n", true);
        check("mn", false);
        check("q", false);
        check("nnnn", false);
    }
    private static void check(String text, boolean expect) {
        boolean result = Pattern.matches("a*|[mn]", text);
        System.out.println(
            (result ? "Match   " : "No match") +
            (result == expect ? " OK    " : " ERROR ") +
            ": " + text
        );
    }
}

...though obviously if you were really using the pattern repeatedly, you'd want to compile it once and reuse the result.

Upvotes: 4

Yasen
Yasen

Reputation: 4474

Try this regex

(a*)|m|n

Pattern.matches("(a*)|m|n", "") // true, match 1st group
Pattern.matches("(a*)|m|n", "a") // true, match 1st group
Pattern.matches("(a*)|m|n", "aaaa") // true, match 1st group
Pattern.matches("(a*)|m|n", "m") // true, match `n`
Pattern.matches("(a*)|m|n", "n") // true, match `m`
Pattern.matches("(a*)|m|n", "man") // false
Pattern.matches("(a*)|m|n", "mn") // false

Upvotes: 1

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