Detecting emoticons with regex in javascript

Question

I am creating a class to detect emoticons and have the following to detect :) smilies including their various variations (=], =), [=, [:, etc.), but it doesn't work and I can't for the life of me figure out what is wrong. I'm testing it in JSFiddle.

var DetectEmoticons = {
    countHappy: function(data) {
        var pattern = new RegExp("[:/=]-?[]/)] | [[/(]-?[:/=]", "g");
        var count = (data.match(pattern) || []).length;
        return count;
    }
}
alert(DetectEmoticons.countHappy("=)"));

stackoverflow.com/questions/4425683/… should help.

Joe Slater
– Joe Slater

2013-06-06 05:10:10 +00:00
Commented Jun 6, 2013 at 5:10 — Joe Slater
– Joe Slater, Commented Jun 6, 2013 at 5:10

Patashu · Accepted Answer · 2013-06-06 05:11:24Z

6

[:=;]-?[)(|\\/\]\[]|[)(|\\/\]\[]-?[:=;]

This looks like an unholy mess from hell until you break it down:

[:=;] matches one : or one = or one ;

[)(|\\/\]\[] matches one ), (, |, \ (backslashed because it is a metacharacter), /, ] (backslashed because it is a metacharacter) or [ (backslashed because it is a metacharacter). (We didn't need to backslash ) or ( because they are not metacharacters inside of a character class).

The | in the center means 'match left of me OR match right of me' and then I write the same two character classes but in reverse to match smileys that are reversed in direction.

I think the problem with your regex is here:

[]/)]

You forgot to escape the first ] with a \ so it is treated as ending the character class prematurely.

The other problem is that you thought forward slash / is used to escape. It's not, \ is used to escape, / has no special meaning in regex.

answered Jun 6, 2013 at 5:11

Patashu

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4 Comments

adrianmcli Over a year ago

Thanks for the explanation, but how come it still doesn't work? jsfiddle.net/TTWsa

Qantas 94 Heavy Over a year ago

You need double the backslashes, i.e. \\ becomes \\\\, and \ becomes \\. This is because of both string and RegExp escaping.

Patashu Over a year ago

@ibopm Do you have to double up backslashes?

Qantas 94 Heavy Over a year ago

If it's going to be static, you can also use a RegExp literal instead of new RegExp(), and avoid the backslash issue. jsfiddle.net/TTWsa/2

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