Question

Good evening, i hope you can help me with this problem, as I’m struggling to find solutions.

I have a provider of words, who give me vowelled Hebrew words , for example –

Vowelled – בַּיִת not vowelled – בית

Vowelled – הַבַּיְתָה not vowelled – הביתה

Unlike my provider, my user can’t normally enter Hebrew vowels (nor should i want him to do that). The user story is the user seeking a word in the provided words. The problem is the comparison between the vowelled and the un-vowelled words. As each is represented by a different byte array in the memory, the equals method returns false.

I tried looking into how UTF-8 handles hebrew vowels and it seems like it’s just normal characters.

I do want to present the vowels to the user, so i want to keep the string as-is in the memory, but when comparing i want to ignore them. Is there any simple way to solve this problem?

Answer 1

You can using a Collator. I can’t tell you how exactly it’s working as it’s new to me, but this appears to do the trick:

public static void main( String[] args ) {
    String withVowels = "בַּיִת";
    String withoutVowels = "בית";

    String withVowelsTwo = "הַבַּיְתָה";
    String withoutVowelsTwo = "הביתה";

    System.out.println( "These two strings are " + (withVowels.equals( withoutVowels ) ? "" : "not ") + "equal" );
    System.out.println( "The second two strings are " + (withVowelsTwo.equals( withoutVowelsTwo ) ? "" : "not ") + "equal" );

    Collator collator = Collator.getInstance( new Locale( "he" ) );
    collator.setStrength( Collator.PRIMARY );

    System.out.println( collator.equals( withVowels, withoutVowels ) );
    System.out.println( collator.equals( withVowelsTwo, withoutVowelsTwo ) );
}

From that, I get the following output:

These two strings are not equal
The second two strings are not equal
true
true