Detecting if a character in a String is an emoticon (using Android)

Issue

Like the title says. I want to find out if a given java String contains an emoticon.

I can’t use Character.UnicodeBlock.of(char) == Character.UnicodeBlock.EMOTICONS since that requires API level 19.

I found this code for iOS but it isn’t really applicable since it looks like java and objective-c handle surrogate pairs in different manners.

The documentations I’ve looked through tell me that:

A char value, therefore, represents Basic Multilingual Plane (BMP) code points, including the surrogate code points, or code units of the UTF-16 encoding

I’m not quite sure what that means. Does that simply mean that they also include the BMP point as their first number?

According to Wikipedia the emoticon set lies between 0x1f600 and 0x1f64f but I don’t know how to check if the char is in that range.

I had hoped that something like this would work but it didn’t

if (0x1f600 <= a && a <= 0x1f64f)
{
    Print.d("Unicode", "groovy!");
}

So how do I go about this?

Solution

I was in fact able to use the linked iOS code to create the following function. I didn’t realize that a String that contains, for example, a single emoticon will have a length of 2. So you can check if a character is in fact a surrogate.

I’m not entirely sure how to handle else if (substring.length > 1) from the iOS code but I think Character.isHighSurrogate(myChar) does the same job in that instance.

private boolean containsIllegalCharacters(String displayName)
{
    final int nameLength = displayName.length();

    for (int i = 0; i < nameLength; i++)
    {
        final char hs = displayName.charAt(i);

        if (0xd800 <= hs && hs <= 0xdbff)
        {
            final char ls = displayName.charAt(i + 1);
            final int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;

            if (0x1d000 <= uc && uc <= 0x1f77f)
            {
                return true;
            }
        }
        else if (Character.isHighSurrogate(hs))
        {
            final char ls = displayName.charAt(i + 1);

            if (ls == 0x20e3)
            {
                return true;
            }
        }
        else
        {
            // non surrogate
            if (0x2100 <= hs && hs <= 0x27ff)
            {
                return true;
            }
            else if (0x2B05 <= hs && hs <= 0x2b07)
            {
                return true;
            }
            else if (0x2934 <= hs && hs <= 0x2935)
            {
                return true;
            }
            else if (0x3297 <= hs && hs <= 0x3299)
            {
                return true;
            }
            else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50)
            {
                return true;
            }
        }
    }

    return false;
}

Answered By – Bjorninn

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

Leave a Reply

(*) Required, Your email will not be published