I have to validate Name fields in my project and have different validation conditions for it.
Starting Name:
Min. length of 2 chars.
Can be alphabetic chars, blanks and hyphens.
The 1st char must be alphabetic.
Blanks & Hyphens must not be adjacent
My Regex:-
function fname(value){
var fn = new RegExp("([a-zA-Z]{1}[a-zA-Z]*[ -]{0,1}[a-zA-Z])+([ -]{0,1}[a-zA-Z]+)*");
if(fn.test(value)){
return true;
}
else{
return false;
}
}
Last Name:
Can be alphabetic chars, blanks, hyphens and apostrophes.
1st char must be alphabetic.
Blanks, hyphens and apostrophes must not be adjacent.
1 char is acceptable only if that char is O
My Regex:
function fname(value){
var ln = new RegExp("([a-zA-Z]+([a-zA-Z]|([ '][a-zA-Z])|([-][a-zA-Z])){1,}|[O]{1})");
if(ln.test(value)){
return true;
}
else{
return false;
}
}
Both these regex are failing as they are accepting:
Alphanumeric characters are getting acceptable which shouldn't be.
space, hyphen adjacent (in case of starting name) and space, hyphen and apostrophes adjacent (in case of last name) at any position in string.
While this can be done with a single regex, the easiest solution is to just separate the tests:
function is_valid_first_name(str) {
return (
str.length >= 2 &&
/^[a-zA-Z \-]*$/.test(str) &&
/^[a-zA-Z]/.test(str) &&
!/[ \-]{2}/.test(str)
);
}
function is_valid_last_name(str) {
return (
/^[a-zA-Z \-']*$/.test(str) &&
/^[a-zA-Z]/.test(str) &&
!/[ \-']{2}/.test(str) &&
(str.length > 1 || str === 'O')
);
}
At first glance I see you aren't specifying start and end of string boundaries in your regexps so any string conatining a matching substring will validate:
> "Hello".match(/[a-z]+/)
[ 'ello', index: 1, input: 'Hello' ]
> "Hello".match(/^[a-z]+$/)
null
Another ugly thing I see are unescaped hyphens in character classes ([ -]). Even in this case is valid (it will work as you expect) but is ugly because it has special meaning in character classes (depending on context) so minimal and apparently inoffensive changes may break things that were already working:
> "-".match(/[abc-]/);
[ '-', index: 0, input: '-' ]
> "-".match(/[abc-j]/);
null
Answering your question: Ignoring 4th rule it's pretty easy:
> "hello-world as foo".match(/^[a-zA-Z][a-zA-Z\s\-]+$/)
[ 'hello-world as foo', index: 0, input: 'hello-world as foo' ]
First character class ([a-zA-Z]) ensures 2nd rule and is compatible with rules n. 1 and 3 too.
The second character class ([a-zA-Z\s\-]) matches any valid character according rule n. 2.
Following quantifier (+) ensures one or more occurrences which, plus the initial character matched by initial character class sums 2 or more characters (so fits rule n. 1).
Finally, starting and ending ^ and $ boundaries ensures, that matching starts from the really beginning of the string and ends at its end so, as I explained earlier, matching substrings aren't enough to validate if invalid characters are present.
4th rule is a bit more tricky. I think it can be approached by lookbehind and lookahead expressions, but they are not available on all regex engines (even in javascript I think they aren't at least in some ancient versions).
...and, even if available, they are always suboptimal (from the regex engine implementation point of view).
On the other hand, you could rely on grouping instead of character classes combining groups with spaces and groups with hyphens, but that will darken your final expression apart of making it harder to build, understand an test. So, in my opinion it isn't a good solution too.
BUT if you are not forced to use single regular expression it's pretty easy to check apart by an ad hoc expression.
function fname(value){
return !!( // <- (optional) Boolean cast for more consistent return type.
value.match(/^[a-zA-Z][a-zA-Z\s\-]+$/)
&& ! value.match(/\s-|-\s/)
);
}
console.log (fname("hello-world as foo")); // true
console.log (fname("hello- world as foo")); // false
console.log (fname("hello -world as foo")); // false
console.log (fname("-hello-world as foo")); // false (null without "!!" cast).
console.log (fname(" hello-world as foo")); // false (null without "!!" cast).
...as a final note, I used "\s" character class instead of "" for spaces. This matches other spacing characters too (like tabs and, in some conditions, line breaks, etc...) if you don't want to accept those characters, replace all "\s" occurrence by simple spaces.
I preferred to use "\s" in the sake of readability (and because in most cases I like it much more if other spacings are acceptable, but I think in this case it's not).
Last name rules are pretty much the same so required changes are trivial following the same reasoning.
First, both functions have the same name. Maybe that's just a typo.
In any case, I think this does what you want for the first name. It doesn't allow a trailing space or hyphen, but that seems to be implied.
const re_fname = /^[a-zA-Z](?:[- ]?|(?:[a-zA-Z][- ]?)+)$/;
function fname(value){
const res = re_fname.test(value);
console.log("%s: %s", res ? "PASS" : "FAIL", value);
return res;
}
fname("foo-bar");
fname("foobar");
fname("f-");
fname("f--");
fname("foo-bar-");
fname("foo-bar--");
fname("-foo-bar");
fname("foo--bar");
fname("foo bar");
fname("foo bar");
fname("foo- bar");
And the last name, which is nearly identical, only adding the apostrophe to the set, and allowing for a single O match.
const re_lname = /^(?:O|(?:[' -]?|[a-zA-Z](?:[' -]?[a-zA-Z])+)[' -]?)$/;
function lname(value){
const res = re_lname.test(value);
console.log("%s: %s", res ? "PASS" : "FAIL", value);
return res;
}
lname("O");
lname("X");
lname("foobar");
lname("foo-bar");
lname("foo-bar-");
lname("foo-bar-'");
lname("foo-bar'");
lname("-foo-bar");
lname("foo--bar");
lname("foo bar");
lname("foo bar");
lname("foo- bar");
lname("foo'bar");
lname("foo' bar");
lname("foo'- bar");
lname("foo-'bar");
Related
I've written a basic 2 operand calculator app (+ - * /) that uses a couple of inline regex validations to filter away invalid characters as they are typed.
An example looks like:
//check if operator is present
if(/[+\-*\/]/.test(display1.textContent)){
//validate the string each time a new character is added
if(!/^\d+\.?\d*[+\-*\/]?\d*\.?\d*$/.test(display1.textContent)){
console.log('invalid')
return false
}
//validate the string character by character before operator
} else {
if(!/^\d+\.?\d*$/.test(display1.textContent)){
console.log('invalid')
return false
}
}
In the above, a valid character doesn't return false:
23.4x0.00025 (no false returned and hence the string is typed out)
But, if an invalid character is typed the function returns false and the input is filtered away:
23.4x0.(x) x at the end returns a false so is filtered (only one operator allowed per calculation)
23.4x0. is typed
It works pretty well but allows for the following which I would like to deal with:
2.+.1
I would prefer 2.0+0.1
My regex would need an if-then-else conditional stating that if the current character is '.' then the next character must be a number else the next char can be number|.|operator. Or if the current character is [+-*/] then the next character must be a number, else the next char can be any char (while following the overall logic).
The tricky part is that the logic must process the string as it is typed character by character and validate at each addition (and be accurate), not at the end when the string is complete.
if-then-else regex is not supported in JavaScript (which I think would satisfy my needs) so I need to use another approach whilst remaining within the JS domain.
Any suggestions about this specific problem would be really helpful.
Thanks
https://github.com/jdineley/Project-calculator
Thanks #trincot for the tips using capturing groups and look around. This helped me write what I needed:
https://regex101.com/r/khUd8H/1
git hub app is updated and works as desired. Now just need to make it pretty!
For ensuring that an operator is not allowed when the preceding number ended in a point, you can insert a positive look behind in your regex that requires the character before an operator to always be a digit: (?<=\d)
Demo:
const validate = s => /^(\d+(\.\d*)?((?<=\d)[+*/-]|$))*$/.test(s);
document.querySelector("input").addEventListener("input", function () {
this.style.backgroundColor = validate(this.value) ? "" : "orange";
});
Input: <input>
I need help for how to detect if an input contains a Japanese emoji/emoticon.
Currently my character set is charset=utf-8. On inputting text, the user can enter Japanese characters/alpanumerics/symbols but if they insert an emoji, onsubmit JavaScript will check if there is an emoji, error message will display.
I can't get this to work because I have no idea on how to detect an emoji in JavaScript?
The answers might work but are terrible because they rely on unicode ranges that are unreadable and somewhat "magic" because it's not always clear where do they come from and why they work, not to mention they're not resilient to new emojis being added to the spec.
Major browsers now support unicode property escape which allows for matching emojis based on their belonging in the Emoji unicode category: \p{Emoji} matches an emoji, \P{Emoji} matches a non-emoji.
Note that officially, 0123456789#* and other characters are emojis too, so the property escape you might want to use is not Emoji but rather Extended_Pictographic which denotes all the characters typically understood as emojis!
Make sure to include the u flag at the end.
console.log(
/\p{Emoji}/u.test('flowers'), // false :)
/\p{Emoji}/u.test('flowers ๐ผ๐บ๐ธ'), // true :)
/\p{Emoji}/u.test('flowers 123'), // true :(
)
console.log(
/\p{Extended_Pictographic}/u.test('flowers'), // false :)
/\p{Extended_Pictographic}/u.test('flowers ๐ผ๐บ๐ธ'), // true :)
/\p{Extended_Pictographic}/u.test('flowers 123'), // false :)
)
This works fine for detecting emojis, but if you want to use the same regex to extract them, you might be surprised with its behavior, since some emojis that appear as one character are actually several characters. They're what we call emoji sequences, more about them in this question
const regex = /\p{Extended_Pictographic}/ug
const family = '๐จโ๐ฉโ๐ง' // "family
console.log(family.length) // not 1, but 8!
console.log(regex.test(family)) // true, as expected
console.log(family.match(regex)) // not [family], but [man, woman, girl]
You can use the following regex:
/(?:[\u2700-\u27bf]|(?:\ud83c[\udde6-\uddff]){2}|[\ud800-\udbff][\udc00-\udfff]|[\u0023-\u0039]\ufe0f?\u20e3|\u3299|\u3297|\u303d|\u3030|\u24c2|\ud83c[\udd70-\udd71]|\ud83c[\udd7e-\udd7f]|\ud83c\udd8e|\ud83c[\udd91-\udd9a]|\ud83c[\udde6-\uddff]|\ud83c[\ude01-\ude02]|\ud83c\ude1a|\ud83c\ude2f|\ud83c[\ude32-\ude3a]|\ud83c[\ude50-\ude51]|\u203c|\u2049|[\u25aa-\u25ab]|\u25b6|\u25c0|[\u25fb-\u25fe]|\u00a9|\u00ae|\u2122|\u2139|\ud83c\udc04|[\u2600-\u26FF]|\u2b05|\u2b06|\u2b07|\u2b1b|\u2b1c|\u2b50|\u2b55|\u231a|\u231b|\u2328|\u23cf|[\u23e9-\u23f3]|[\u23f8-\u23fa]|\ud83c\udccf|\u2934|\u2935|[\u2190-\u21ff])/g
If you just want to remove it from the string, you can do something like this.
function removeEmojis (string) {
var regex = /(?:[\u2700-\u27bf]|(?:\ud83c[\udde6-\uddff]){2}|[\ud800-\udbff][\udc00-\udfff]|[\u0023-\u0039]\ufe0f?\u20e3|\u3299|\u3297|\u303d|\u3030|\u24c2|\ud83c[\udd70-\udd71]|\ud83c[\udd7e-\udd7f]|\ud83c\udd8e|\ud83c[\udd91-\udd9a]|\ud83c[\udde6-\uddff]|\ud83c[\ude01-\ude02]|\ud83c\ude1a|\ud83c\ude2f|\ud83c[\ude32-\ude3a]|\ud83c[\ude50-\ude51]|\u203c|\u2049|[\u25aa-\u25ab]|\u25b6|\u25c0|[\u25fb-\u25fe]|\u00a9|\u00ae|\u2122|\u2139|\ud83c\udc04|[\u2600-\u26FF]|\u2b05|\u2b06|\u2b07|\u2b1b|\u2b1c|\u2b50|\u2b55|\u231a|\u231b|\u2328|\u23cf|[\u23e9-\u23f3]|[\u23f8-\u23fa]|\ud83c\udccf|\u2934|\u2935|[\u2190-\u21ff])/g;
return string.replace(regex, '');
}
A simple function that returns true if your string contains one or more emojis.
function isEmoji(str) {
var ranges = [
'(?:[\u2700-\u27bf]|(?:\ud83c[\udde6-\uddff]){2}|[\ud800-\udbff][\udc00-\udfff]|[\u0023-\u0039]\ufe0f?\u20e3|\u3299|\u3297|\u303d|\u3030|\u24c2|\ud83c[\udd70-\udd71]|\ud83c[\udd7e-\udd7f]|\ud83c\udd8e|\ud83c[\udd91-\udd9a]|\ud83c[\udde6-\uddff]|[\ud83c[\ude01-\ude02]|\ud83c\ude1a|\ud83c\ude2f|[\ud83c[\ude32-\ude3a]|[\ud83c[\ude50-\ude51]|\u203c|\u2049|[\u25aa-\u25ab]|\u25b6|\u25c0|[\u25fb-\u25fe]|\u00a9|\u00ae|\u2122|\u2139|\ud83c\udc04|[\u2600-\u26FF]|\u2b05|\u2b06|\u2b07|\u2b1b|\u2b1c|\u2b50|\u2b55|\u231a|\u231b|\u2328|\u23cf|[\u23e9-\u23f3]|[\u23f8-\u23fa]|\ud83c\udccf|\u2934|\u2935|[\u2190-\u21ff])' // U+1F680 to U+1F6FF
];
if (str.match(ranges.join('|'))) {
return true;
} else {
return false;
}
}
First of all, you cannot rely on ECMAScript 2018+ compliant \p{Emoji} (at least at the time of writing). It really matches some 0123456789#* non-emoji chars (see Nino Filiu's answer). See Why do Unicode emoji property escapes match numbers? for more details.
To test if there are any emoji chars in a string in JavaScript, you can use the following ECMAScript 2018+ compliant solution (mind the u flag):
const regex_emoji = /[\p{Extended_Pictographic}\u{1F3FB}-\u{1F3FF}\u{1F9B0}-\u{1F9B3}]/u;
console.log( regex_emoji.test('flowers 123') ); // => false
console.log( regex_emoji.test('flowers ๐ผ๐บ๐ธ') ); // => true
You can even extract one or more emoji char sequences using this pattern (note the added g flag to find all occurrences and + to match one or more consecutive occurrences of the character class pattern):
const regex_emoji = /[\p{Extended_Pictographic}\u{1F3FB}-\u{1F3FF}\u{1F9B0}-\u{1F9B3}]+/gu;
console.log( 'flowers 123'.match(regex_emoji) ); // => null
console.log( 'flowers ๐ผ๐บ๐ธ'.match(regex_emoji) ); // => [ "๐ผ๐บ๐ธ" ]
In a nutshell, the Extended_Pictographic Unicode category class matches most emoji chars except for some Emoji_Components, that is, light skin to dark skin mode chars (\u{1F3FB}-\u{1F3FF}) and red-haired to white-haired chars (\u{1F9B0}-\u{1F9B3}).
To count, or extract emojis as an array of single chars or consecutive sequences (like in the second code snippet from above) from longer texts and perform other actions on emojis, you can use a custom regex (like in Scott Weaver's answer). It is safer to use the longer, escaped version. However, there are 4702 emoji characters defined in the Emoji Keyboard/Display Test Data for UTR #51 (Version: 14.0) file. Thus, the (ES5 compliant, works even in IE) regex to match a single emoji char is
var EmojiPattern = /[#*0-9]\uFE0F?\u20E3|\u00A9\uFE0F?|[\u00AE\u203C\u2049\u2122\u2139\u2194-\u2199\u21A9\u21AA]\uFE0F?|[\u231A\u231B]|[\u2328\u23CF]\uFE0F?|[\u23E9-\u23EC]|[\u23ED-\u23EF]\uFE0F?|\u23F0|[\u23F1\u23F2]\uFE0F?|\u23F3|[\u23F8-\u23FA\u24C2\u25AA\u25AB\u25B6\u25C0\u25FB\u25FC]\uFE0F?|[\u25FD\u25FE]|[\u2600-\u2604\u260E\u2611]\uFE0F?|[\u2614\u2615]|\u2618\uFE0F?|\u261D(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F)?|[\u2620\u2622\u2623\u2626\u262A\u262E\u262F\u2638-\u263A\u2640\u2642]\uFE0F?|[\u2648-\u2653]|[\u265F\u2660\u2663\u2665\u2666\u2668\u267B\u267E]\uFE0F?|\u267F|\u2692\uFE0F?|\u2693|[\u2694-\u2697\u2699\u269B\u269C\u26A0]\uFE0F?|\u26A1|\u26A7\uFE0F?|[\u26AA\u26AB]|[\u26B0\u26B1]\uFE0F?|[\u26BD\u26BE\u26C4\u26C5]|\u26C8\uFE0F?|\u26CE|[\u26CF\u26D1\u26D3]\uFE0F?|\u26D4|\u26E9\uFE0F?|\u26EA|[\u26F0\u26F1]\uFE0F?|[\u26F2\u26F3]|\u26F4\uFE0F?|\u26F5|[\u26F7\u26F8]\uFE0F?|\u26F9(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?|\uFE0F(?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\u26FA\u26FD]|\u2702\uFE0F?|\u2705|[\u2708\u2709]\uFE0F?|[\u270A\u270B](?:\uD83C[\uDFFB-\uDFFF])?|[\u270C\u270D](?:\uD83C[\uDFFB-\uDFFF]|\uFE0F)?|\u270F\uFE0F?|[\u2712\u2714\u2716\u271D\u2721]\uFE0F?|\u2728|[\u2733\u2734\u2744\u2747]\uFE0F?|[\u274C\u274E\u2753-\u2755\u2757]|\u2763\uFE0F?|\u2764(?:\u200D(?:\uD83D\uDD25|\uD83E\uDE79)|\uFE0F(?:\u200D(?:\uD83D\uDD25|\uD83E\uDE79))?)?|[\u2795-\u2797]|\u27A1\uFE0F?|[\u27B0\u27BF]|[\u2934\u2935\u2B05-\u2B07]\uFE0F?|[\u2B1B\u2B1C\u2B50\u2B55]|[\u3030\u303D\u3297\u3299]\uFE0F?|\uD83C(?:[\uDC04\uDCCF]|[\uDD70\uDD71\uDD7E\uDD7F]\uFE0F?|[\uDD8E\uDD91-\uDD9A]|\uDDE6\uD83C[\uDDE8-\uDDEC\uDDEE\uDDF1\uDDF2\uDDF4\uDDF6-\uDDFA\uDDFC\uDDFD\uDDFF]|\uDDE7\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEF\uDDF1-\uDDF4\uDDF6-\uDDF9\uDDFB\uDDFC\uDDFE\uDDFF]|\uDDE8\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDEE\uDDF0-\uDDF5\uDDF7\uDDFA-\uDDFF]|\uDDE9\uD83C[\uDDEA\uDDEC\uDDEF\uDDF0\uDDF2\uDDF4\uDDFF]|\uDDEA\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDED\uDDF7-\uDDFA]|\uDDEB\uD83C[\uDDEE-\uDDF0\uDDF2\uDDF4\uDDF7]|\uDDEC\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEE\uDDF1-\uDDF3\uDDF5-\uDDFA\uDDFC\uDDFE]|\uDDED\uD83C[\uDDF0\uDDF2\uDDF3\uDDF7\uDDF9\uDDFA]|\uDDEE\uD83C[\uDDE8-\uDDEA\uDDF1-\uDDF4\uDDF6-\uDDF9]|\uDDEF\uD83C[\uDDEA\uDDF2\uDDF4\uDDF5]|\uDDF0\uD83C[\uDDEA\uDDEC-\uDDEE\uDDF2\uDDF3\uDDF5\uDDF7\uDDFC\uDDFE\uDDFF]|\uDDF1\uD83C[\uDDE6-\uDDE8\uDDEE\uDDF0\uDDF7-\uDDFB\uDDFE]|\uDDF2\uD83C[\uDDE6\uDDE8-\uDDED\uDDF0-\uDDFF]|\uDDF3\uD83C[\uDDE6\uDDE8\uDDEA-\uDDEC\uDDEE\uDDF1\uDDF4\uDDF5\uDDF7\uDDFA\uDDFF]|\uDDF4\uD83C\uDDF2|\uDDF5\uD83C[\uDDE6\uDDEA-\uDDED\uDDF0-\uDDF3\uDDF7-\uDDF9\uDDFC\uDDFE]|\uDDF6\uD83C\uDDE6|\uDDF7\uD83C[\uDDEA\uDDF4\uDDF8\uDDFA\uDDFC]|\uDDF8\uD83C[\uDDE6-\uDDEA\uDDEC-\uDDF4\uDDF7-\uDDF9\uDDFB\uDDFD-\uDDFF]|\uDDF9\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDED\uDDEF-\uDDF4\uDDF7\uDDF9\uDDFB\uDDFC\uDDFF]|\uDDFA\uD83C[\uDDE6\uDDEC\uDDF2\uDDF3\uDDF8\uDDFE\uDDFF]|\uDDFB\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDEE\uDDF3\uDDFA]|\uDDFC\uD83C[\uDDEB\uDDF8]|\uDDFD\uD83C\uDDF0|\uDDFE\uD83C[\uDDEA\uDDF9]|\uDDFF\uD83C[\uDDE6\uDDF2\uDDFC]|\uDE01|\uDE02\uFE0F?|[\uDE1A\uDE2F\uDE32-\uDE36]|\uDE37\uFE0F?|[\uDE38-\uDE3A\uDE50\uDE51\uDF00-\uDF20]|[\uDF21\uDF24-\uDF2C]\uFE0F?|[\uDF2D-\uDF35]|\uDF36\uFE0F?|[\uDF37-\uDF7C]|\uDF7D\uFE0F?|[\uDF7E-\uDF84]|\uDF85(?:\uD83C[\uDFFB-\uDFFF])?|[\uDF86-\uDF93]|[\uDF96\uDF97\uDF99-\uDF9B\uDF9E\uDF9F]\uFE0F?|[\uDFA0-\uDFC1]|\uDFC2(?:\uD83C[\uDFFB-\uDFFF])?|[\uDFC3\uDFC4](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDFC5\uDFC6]|\uDFC7(?:\uD83C[\uDFFB-\uDFFF])?|[\uDFC8\uDFC9]|\uDFCA(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDFCB\uDFCC](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?|\uFE0F(?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDFCD\uDFCE]\uFE0F?|[\uDFCF-\uDFD3]|[\uDFD4-\uDFDF]\uFE0F?|[\uDFE0-\uDFF0]|\uDFF3(?:\u200D(?:\u26A7\uFE0F?|\uD83C\uDF08)|\uFE0F(?:\u200D(?:\u26A7\uFE0F?|\uD83C\uDF08))?)?|\uDFF4(?:\u200D\u2620\uFE0F?|\uDB40\uDC67\uDB40\uDC62\uDB40(?:\uDC65\uDB40\uDC6E\uDB40\uDC67|\uDC73\uDB40\uDC63\uDB40\uDC74|\uDC77\uDB40\uDC6C\uDB40\uDC73)\uDB40\uDC7F)?|[\uDFF5\uDFF7]\uFE0F?|[\uDFF8-\uDFFF])|\uD83D(?:[\uDC00-\uDC07]|\uDC08(?:\u200D\u2B1B)?|[\uDC09-\uDC14]|\uDC15(?:\u200D\uD83E\uDDBA)?|[\uDC16-\uDC3A]|\uDC3B(?:\u200D\u2744\uFE0F?)?|[\uDC3C-\uDC3E]|\uDC3F\uFE0F?|\uDC40|\uDC41(?:\u200D\uD83D\uDDE8\uFE0F?|\uFE0F(?:\u200D\uD83D\uDDE8\uFE0F?)?)?|[\uDC42\uDC43](?:\uD83C[\uDFFB-\uDFFF])?|[\uDC44\uDC45]|[\uDC46-\uDC50](?:\uD83C[\uDFFB-\uDFFF])?|[\uDC51-\uDC65]|[\uDC66\uDC67](?:\uD83C[\uDFFB-\uDFFF])?|\uDC68(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?|[\uDC68\uDC69]\u200D\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?)|[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92])|\uD83E[\uDDAF-\uDDB3\uDDBC\uDDBD])|\uD83C(?:\uDFFB(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68\uD83C[\uDFFB-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D\uDC68\uD83C[\uDFFC-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFC(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68\uD83C[\uDFFB-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D\uDC68\uD83C[\uDFFB\uDFFD-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFD(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68\uD83C[\uDFFB-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D\uDC68\uD83C[\uDFFB\uDFFC\uDFFE\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFE(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68\uD83C[\uDFFB-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D\uDC68\uD83C[\uDFFB-\uDFFD\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFF(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?\uDC68\uD83C[\uDFFB-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D\uDC68\uD83C[\uDFFB-\uDFFE]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?))?|\uDC69(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:\uDC8B\u200D\uD83D)?[\uDC68\uDC69]|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?|\uDC69\u200D\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?)|[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92])|\uD83E[\uDDAF-\uDDB3\uDDBC\uDDBD])|\uD83C(?:\uDFFB(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF]|\uDC8B\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF])|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFC-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFC(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF]|\uDC8B\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF])|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB\uDFFD-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFD(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF]|\uDC8B\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF])|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB\uDFFC\uDFFE\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFE(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF]|\uDC8B\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF])|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFD\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFF(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D\uD83D(?:[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF]|\uDC8B\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFF])|\uD83C[\uDF3E\uDF73\uDF7C\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83D[\uDC68\uDC69]\uD83C[\uDFFB-\uDFFE]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?))?|\uDC6A|[\uDC6B-\uDC6D](?:\uD83C[\uDFFB-\uDFFF])?|\uDC6E(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDC6F(?:\u200D[\u2640\u2642]\uFE0F?)?|[\uDC70\uDC71](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDC72(?:\uD83C[\uDFFB-\uDFFF])?|\uDC73(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDC74-\uDC76](?:\uD83C[\uDFFB-\uDFFF])?|\uDC77(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDC78(?:\uD83C[\uDFFB-\uDFFF])?|[\uDC79-\uDC7B]|\uDC7C(?:\uD83C[\uDFFB-\uDFFF])?|[\uDC7D-\uDC80]|[\uDC81\uDC82](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDC83(?:\uD83C[\uDFFB-\uDFFF])?|\uDC84|\uDC85(?:\uD83C[\uDFFB-\uDFFF])?|[\uDC86\uDC87](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDC88-\uDC8E]|\uDC8F(?:\uD83C[\uDFFB-\uDFFF])?|\uDC90|\uDC91(?:\uD83C[\uDFFB-\uDFFF])?|[\uDC92-\uDCA9]|\uDCAA(?:\uD83C[\uDFFB-\uDFFF])?|[\uDCAB-\uDCFC]|\uDCFD\uFE0F?|[\uDCFF-\uDD3D]|[\uDD49\uDD4A]\uFE0F?|[\uDD4B-\uDD4E\uDD50-\uDD67]|[\uDD6F\uDD70\uDD73]\uFE0F?|\uDD74(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F)?|\uDD75(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?|\uFE0F(?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDD76-\uDD79]\uFE0F?|\uDD7A(?:\uD83C[\uDFFB-\uDFFF])?|[\uDD87\uDD8A-\uDD8D]\uFE0F?|\uDD90(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F)?|[\uDD95\uDD96](?:\uD83C[\uDFFB-\uDFFF])?|\uDDA4|[\uDDA5\uDDA8\uDDB1\uDDB2\uDDBC\uDDC2-\uDDC4\uDDD1-\uDDD3\uDDDC-\uDDDE\uDDE1\uDDE3\uDDE8\uDDEF\uDDF3\uDDFA]\uFE0F?|[\uDDFB-\uDE2D]|\uDE2E(?:\u200D\uD83D\uDCA8)?|[\uDE2F-\uDE34]|\uDE35(?:\u200D\uD83D\uDCAB)?|\uDE36(?:\u200D\uD83C\uDF2B\uFE0F?)?|[\uDE37-\uDE44]|[\uDE45-\uDE47](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDE48-\uDE4A]|\uDE4B(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDE4C(?:\uD83C[\uDFFB-\uDFFF])?|[\uDE4D\uDE4E](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDE4F(?:\uD83C[\uDFFB-\uDFFF])?|[\uDE80-\uDEA2]|\uDEA3(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDEA4-\uDEB3]|[\uDEB4-\uDEB6](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDEB7-\uDEBF]|\uDEC0(?:\uD83C[\uDFFB-\uDFFF])?|[\uDEC1-\uDEC5]|\uDECB\uFE0F?|\uDECC(?:\uD83C[\uDFFB-\uDFFF])?|[\uDECD-\uDECF]\uFE0F?|[\uDED0-\uDED2\uDED5-\uDED7\uDEDD-\uDEDF]|[\uDEE0-\uDEE5\uDEE9]\uFE0F?|[\uDEEB\uDEEC]|[\uDEF0\uDEF3]\uFE0F?|[\uDEF4-\uDEFC\uDFE0-\uDFEB\uDFF0])|\uD83E(?:\uDD0C(?:\uD83C[\uDFFB-\uDFFF])?|[\uDD0D\uDD0E]|\uDD0F(?:\uD83C[\uDFFB-\uDFFF])?|[\uDD10-\uDD17]|[\uDD18-\uDD1F](?:\uD83C[\uDFFB-\uDFFF])?|[\uDD20-\uDD25]|\uDD26(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDD27-\uDD2F]|[\uDD30-\uDD34](?:\uD83C[\uDFFB-\uDFFF])?|\uDD35(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDD36(?:\uD83C[\uDFFB-\uDFFF])?|[\uDD37-\uDD39](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDD3A|\uDD3C(?:\u200D[\u2640\u2642]\uFE0F?)?|[\uDD3D\uDD3E](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDD3F-\uDD45\uDD47-\uDD76]|\uDD77(?:\uD83C[\uDFFB-\uDFFF])?|[\uDD78-\uDDB4]|[\uDDB5\uDDB6](?:\uD83C[\uDFFB-\uDFFF])?|\uDDB7|[\uDDB8\uDDB9](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDDBA|\uDDBB(?:\uD83C[\uDFFB-\uDFFF])?|[\uDDBC-\uDDCC]|[\uDDCD-\uDDCF](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDDD0|\uDDD1(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1|[\uDDAF-\uDDB3\uDDBC\uDDBD]))|\uD83C(?:\uDFFB(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D(?:\uD83D\uDC8B\u200D)?\uD83E\uDDD1\uD83C[\uDFFC-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1\uD83C[\uDFFB-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFC(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D(?:\uD83D\uDC8B\u200D)?\uD83E\uDDD1\uD83C[\uDFFB\uDFFD-\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1\uD83C[\uDFFB-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFD(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D(?:\uD83D\uDC8B\u200D)?\uD83E\uDDD1\uD83C[\uDFFB\uDFFC\uDFFE\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1\uD83C[\uDFFB-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFE(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D(?:\uD83D\uDC8B\u200D)?\uD83E\uDDD1\uD83C[\uDFFB-\uDFFD\uDFFF]|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1\uD83C[\uDFFB-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?|\uDFFF(?:\u200D(?:[\u2695\u2696\u2708]\uFE0F?|\u2764\uFE0F?\u200D(?:\uD83D\uDC8B\u200D)?\uD83E\uDDD1\uD83C[\uDFFB-\uDFFE]|\uD83C[\uDF3E\uDF73\uDF7C\uDF84\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E(?:\uDD1D\u200D\uD83E\uDDD1\uD83C[\uDFFB-\uDFFF]|[\uDDAF-\uDDB3\uDDBC\uDDBD])))?))?|[\uDDD2\uDDD3](?:\uD83C[\uDFFB-\uDFFF])?|\uDDD4(?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|\uDDD5(?:\uD83C[\uDFFB-\uDFFF])?|[\uDDD6-\uDDDD](?:\u200D[\u2640\u2642]\uFE0F?|\uD83C[\uDFFB-\uDFFF](?:\u200D[\u2640\u2642]\uFE0F?)?)?|[\uDDDE\uDDDF](?:\u200D[\u2640\u2642]\uFE0F?)?|[\uDDE0-\uDDFF\uDE70-\uDE74\uDE78-\uDE7C\uDE80-\uDE86\uDE90-\uDEAC\uDEB0-\uDEBA\uDEC0-\uDEC2]|[\uDEC3-\uDEC5](?:\uD83C[\uDFFB-\uDFFF])?|[\uDED0-\uDED9\uDEE0-\uDEE7]|\uDEF0(?:\uD83C[\uDFFB-\uDFFF])?|\uDEF1(?:\uD83C(?:\uDFFB(?:\u200D\uD83E\uDEF2\uD83C[\uDFFC-\uDFFF])?|\uDFFC(?:\u200D\uD83E\uDEF2\uD83C[\uDFFB\uDFFD-\uDFFF])?|\uDFFD(?:\u200D\uD83E\uDEF2\uD83C[\uDFFB\uDFFC\uDFFE\uDFFF])?|\uDFFE(?:\u200D\uD83E\uDEF2\uD83C[\uDFFB-\uDFFD\uDFFF])?|\uDFFF(?:\u200D\uD83E\uDEF2\uD83C[\uDFFB-\uDFFE])?))?|[\uDEF2-\uDEF6](?:\uD83C[\uDFFB-\uDFFF])?)/g;
See the regex demo. The g flag at the end means this regex can match all occurrences in the input string.
The pattern is created dynamically from the list of emojis and contracted using a regex trie.
See this JavaScript demo:
var text = 'flowers ๐ผ๐บ๐ธ';
// Detecting if there is at least one emoji
console.log( emoji_detection_regex.test(text) ); // => true
// Counting emojis
console.log( (text.match(emoji_count_regex) || []).length ); // => 3
// Extracting one by one, single emoji array
console.log( text.match(emoji_count_regex) ); // => ["๐ผ","๐บ","๐ธ"]
// Extracting emoji sequences
console.log( text.match(emoji_extract_or_remove_regex) ); // => ["๐ผ๐บ๐ธ"]
// Removing emojis
console.log( text.replace(emoji_extract_or_remove_regex, '') ); // => 'flowers '
<script src="https://gitcdn.link/repo/stribizhev/Emojis/main/ws_emoji_regex.js"></script>
The emoji regex declarations are available in the https://github.com/stribizhev/Emojis/blob/main/ws_emoji_regex.js file.
We can detect all list of surrogate pairs or the Emoji characters in a specific range.
If the issue related with storing the input string to database like MySQL version before 5.5 we need to detect and remove all the surrogate pairs using the below regex
/([\uD800-\uDBFF][\uDC00-\uDFFF])/g.
Update for 2020: Many of these patterns don't match compound emojis or Modifier Sequences correctly, or are simply outdated and do not match the newer emojis.
Consider this kissing couple: ๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ. It's actually 6 (maybe more) other emojis glued together with the ZWJ zero-width joiner. To match this correctly, you have to actually match that sequence.
Thus, by matching the longer sequences first, this brute-force pattern (too long too paste, but it's a simple alternation and runs fast) correctly parses all 3521 combined emojis as of May 2021:
GitHub link: https://github.com/sweaver2112/Regex-combined-emojis
Edit 5/10/2021: If the size of the Unicode escape version of this regex is off-putting to you, you could actually skip the Unicode escape sequences and just use literal emojis, thereby saving tons of space...well, almost. This character, "*๏ธโฃ", which starts with an asterisk, will throw a "nothing to quantify" error. Getting rid of just this guy yields a much shorter, still working, copy/pastable regex that matches 3,520/3,521 Emojis at the present date:
Regex 101 Demo (compact, unsafe literal emoji version)
Regex 101 Demo (long, safe unicode escape version)
The demos' input string includes all characters from
https://unicode.org/emoji/charts/full-emoji-list.html (13.1)
https://unicode.org/emoji/charts-13.1/full-emoji-modifiers.html
Working example using the compact version:
/*the pattern*/
var emojiPattern = String.raw`(?:๐ง๐ปโโค๏ธโ๐โ๐ง๐ผ|๐ง๐ปโโค๏ธโ๐โ๐ง๐ฝ|๐ง๐ปโโค๏ธโ๐โ๐ง๐พ|๐ง๐ปโโค๏ธโ๐โ๐ง๐ฟ|๐ง๐ผโโค๏ธโ๐โ๐ง๐ป|๐ง๐ผโโค๏ธโ๐โ๐ง๐ฝ|๐ง๐ผโโค๏ธโ๐โ๐ง๐พ|๐ง๐ผโโค๏ธโ๐โ๐ง๐ฟ|๐ง๐ฝโโค๏ธโ๐โ๐ง๐ป|๐ง๐ฝโโค๏ธโ๐โ๐ง๐ผ|๐ง๐ฝโโค๏ธโ๐โ๐ง๐พ|๐ง๐ฝโโค๏ธโ๐โ๐ง๐ฟ|๐ง๐พโโค๏ธโ๐โ๐ง๐ป|๐ง๐พโโค๏ธโ๐โ๐ง๐ผ|๐ง๐พโโค๏ธโ๐โ๐ง๐ฝ|๐ง๐พโโค๏ธโ๐โ๐ง๐ฟ|๐ง๐ฟโโค๏ธโ๐โ๐ง๐ป|๐ง๐ฟโโค๏ธโ๐โ๐ง๐ผ|๐ง๐ฟโโค๏ธโ๐โ๐ง๐ฝ|๐ง๐ฟโโค๏ธโ๐โ๐ง๐พ|๐ฉ๐ปโโค๏ธโ๐โ๐จ๐ป|๐ฉ๐ปโโค๏ธโ๐โ๐จ๐ผ|๐ฉ๐ปโโค๏ธโ๐โ๐จ๐ฝ|๐ฉ๐ปโโค๏ธโ๐โ๐จ๐พ|๐ฉ๐ปโโค๏ธโ๐โ๐จ๐ฟ|๐ฉ๐ผโโค๏ธโ๐โ๐จ๐ป|๐ฉ๐ผโโค๏ธโ๐โ๐จ๐ผ|๐ฉ๐ผโโค๏ธโ๐โ๐จ๐ฝ|๐ฉ๐ผโโค๏ธโ๐โ๐จ๐พ|๐ฉ๐ผโโค๏ธโ๐โ๐จ๐ฟ|๐ฉ๐ฝโโค๏ธโ๐โ๐จ๐ป|๐ฉ๐ฝโโค๏ธโ๐โ๐จ๐ผ|๐ฉ๐ฝโโค๏ธโ๐โ๐จ๐ฝ|๐ฉ๐ฝโโค๏ธโ๐โ๐จ๐พ|๐ฉ๐ฝโโค๏ธโ๐โ๐จ๐ฟ|๐ฉ๐พโโค๏ธโ๐โ๐จ๐ป|๐ฉ๐พโโค๏ธโ๐โ๐จ๐ผ|๐ฉ๐พโโค๏ธโ๐โ๐จ๐ฝ|๐ฉ๐พโโค๏ธโ๐โ๐จ๐พ|๐ฉ๐พโโค๏ธโ๐โ๐จ๐ฟ|๐ฉ๐ฟโโค๏ธโ๐โ๐จ๐ป|๐ฉ๐ฟโโค๏ธโ๐โ๐จ๐ผ|๐ฉ๐ฟโโค๏ธโ๐โ๐จ๐ฝ|๐ฉ๐ฟโโค๏ธโ๐โ๐จ๐พ|๐ฉ๐ฟโโค๏ธโ๐โ๐จ๐ฟ|๐จ๐ปโโค๏ธโ๐โ๐จ๐ป|๐จ๐ปโโค๏ธโ๐โ๐จ๐ผ|๐จ๐ปโโค๏ธโ๐โ๐จ๐ฝ|๐จ๐ปโโค๏ธโ๐โ๐จ๐พ|๐จ๐ปโโค๏ธโ๐โ๐จ๐ฟ|๐จ๐ผโโค๏ธโ๐โ๐จ๐ป|๐จ๐ผโโค๏ธโ๐โ๐จ๐ผ|๐จ๐ผโโค๏ธโ๐โ๐จ๐ฝ|๐จ๐ผโโค๏ธโ๐โ๐จ๐พ|๐จ๐ผโโค๏ธโ๐โ๐จ๐ฟ|๐จ๐ฝโโค๏ธโ๐โ๐จ๐ป|๐จ๐ฝโโค๏ธโ๐โ๐จ๐ผ|๐จ๐ฝโโค๏ธโ๐โ๐จ๐ฝ|๐จ๐ฝโโค๏ธโ๐โ๐จ๐พ|๐จ๐ฝโโค๏ธโ๐โ๐จ๐ฟ|๐จ๐พโโค๏ธโ๐โ๐จ๐ป|๐จ๐พโโค๏ธโ๐โ๐จ๐ผ|๐จ๐พโโค๏ธโ๐โ๐จ๐ฝ|๐จ๐พโโค๏ธโ๐โ๐จ๐พ|๐จ๐พโโค๏ธโ๐โ๐จ๐ฟ|๐จ๐ฟโโค๏ธโ๐โ๐จ๐ป|๐จ๐ฟโโค๏ธโ๐โ๐จ๐ผ|๐จ๐ฟโโค๏ธโ๐โ๐จ๐ฝ|๐จ๐ฟโโค๏ธโ๐โ๐จ๐พ|๐จ๐ฟโโค๏ธโ๐โ๐จ๐ฟ|๐ฉ๐ปโโค๏ธโ๐โ๐ฉ๐ป|๐ฉ๐ปโโค๏ธโ๐โ๐ฉ๐ผ|๐ฉ๐ปโโค๏ธโ๐โ๐ฉ๐ฝ|๐ฉ๐ปโโค๏ธโ๐โ๐ฉ๐พ|๐ฉ๐ปโโค๏ธโ๐โ๐ฉ๐ฟ|๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ป|๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ผ|๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ|๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐พ|๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฟ|๐ฉ๐ฝโโค๏ธโ๐โ๐ฉ๐ป|๐ฉ๐ฝโโค๏ธโ๐โ๐ฉ๐ผ|๐ฉ๐ฝโโค๏ธโ๐โ๐ฉ๐ฝ|๐ฉ๐ฝโโค๏ธโ๐โ๐ฉ๐พ|๐ฉ๐ฝโโค๏ธโ๐โ๐ฉ๐ฟ|๐ฉ๐พโโค๏ธโ๐โ๐ฉ๐ป|๐ฉ๐พโโค๏ธโ๐โ๐ฉ๐ผ|๐ฉ๐พโโค๏ธโ๐โ๐ฉ๐ฝ|๐ฉ๐พโโค๏ธโ๐โ๐ฉ๐พ|๐ฉ๐พโโค๏ธโ๐โ๐ฉ๐ฟ|๐ฉ๐ฟโโค๏ธโ๐โ๐ฉ๐ป|๐ฉ๐ฟโโค๏ธโ๐โ๐ฉ๐ผ|๐ฉ๐ฟโโค๏ธโ๐โ๐ฉ๐ฝ|๐ฉ๐ฟโโค๏ธโ๐โ๐ฉ๐พ|๐ฉ๐ฟโโค๏ธโ๐โ๐ฉ๐ฟ|๐ด๓ ง๓ ข๓ ฅ๓ ฎ๓ ง๓ ฟ|๐ด๓ ง๓ ข๓ ณ๓ ฃ๓ ด๓ ฟ|๐ด๓ ง๓ ข๓ ท๓ ฌ๓ ณ๓ 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๐ฝโโ๏ธ|๐
๐พโโ๏ธ|๐
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โ๏ธ|๐ง๐พโโ๏ธ|๐ง๐ฟโโ๏ธ|๐ง๐ปโโ๏ธ|๐ง๐ผโโ๏ธ|๐ง๐ฝโโ๏ธ|๐ง๐พโโ๏ธ|๐ง๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐ฃ๐ปโโ๏ธ|๐ฃ๐ผโโ๏ธ|๐ฃ๐ฝโโ๏ธ|๐ฃ๐พโโ๏ธ|๐ฃ๐ฟโโ๏ธ|๐ฃ๐ปโโ๏ธ|๐ฃ๐ผโโ๏ธ|๐ฃ๐ฝโโ๏ธ|๐ฃ๐พโโ๏ธ|๐ฃ๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐๐ปโโ๏ธ|๐๐ผโโ๏ธ|๐๐ฝโโ๏ธ|๐๐พโโ๏ธ|๐๐ฟโโ๏ธ|๐ด๐ปโโ๏ธ|๐ด๐ผโโ๏ธ|๐ด๐ฝโโ๏ธ|๐ด๐พโโ๏ธ|๐ด๐ฟโโ๏ธ|๐ด๐ปโโ๏ธ|๐ด๐ผโโ๏ธ|๐ด๐ฝโโ๏ธ|๐ด๐พโโ๏ธ|๐ด๐ฟโโ๏ธ|๐ต๐ปโโ๏ธ|๐ต๐ผโโ๏ธ|๐ต๐ฝโโ๏ธ|๐ต๐พโโ๏ธ|๐ต๐ฟโโ๏ธ|๐ต๐ปโโ๏ธ|๐ต๐ผโโ๏ธ|๐ต๐ฝโโ๏ธ|๐ต๐พโโ๏ธ|๐ต๐ฟโโ๏ธ|๐คธ๐ปโโ๏ธ|๐คธ๐ผโโ๏ธ|๐คธ๐ฝโโ๏ธ|๐คธ๐พโโ๏ธ|๐คธ๐ฟโโ๏ธ|๐คธ๐ปโโ๏ธ|๐คธ๐ผโโ๏ธ|๐คธ๐ฝโโ๏ธ|๐คธ๐พโโ๏ธ|๐คธ๐ฟโโ๏ธ|๐คฝ๐ปโโ๏ธ|๐คฝ๐ผโโ๏ธ|๐คฝ๐ฝโโ๏ธ|๐คฝ๐พโโ๏ธ|๐คฝ๐ฟโโ๏ธ|๐คฝ๐ปโโ๏ธ|๐คฝ๐ผโโ๏ธ|๐คฝ๐ฝโโ๏ธ|๐คฝ๐พโโ๏ธ|๐คฝ๐ฟโโ๏ธ|๐คพ๐ปโโ๏ธ|๐คพ๐ผโโ๏ธ|๐คพ๐ฝโโ๏ธ|๐คพ๐พโโ๏ธ|๐คพ๐ฟโโ๏ธ|๐คพ๐ปโโ๏ธ|๐คพ๐ผโโ๏ธ|๐คพ๐ฝโโ๏ธ|๐คพ๐พโโ๏ธ|๐คพ๐ฟโโ๏ธ|๐คน๐ปโโ๏ธ|๐คน๐ผโโ๏ธ|๐คน๐ฝโโ๏ธ|๐คน๐พโโ๏ธ|๐คน๐ฟโโ๏ธ|๐คน๐ปโโ๏ธ|๐คน๐ผโโ๏ธ|๐คน๐ฝโโ๏ธ|๐คน๐พโโ๏ธ|๐คน๐ฟโโ๏ธ|๐ง๐ปโโ๏ธ|๐ง๐ผโโ๏ธ|๐ง๐ฝโโ๏ธ|๐ง๐พโโ๏ธ|๐ง๐ฟโโ๏ธ|๐ง๐ปโโ๏ธ|๐ง๐ผโโ๏ธ|๐ง๐ฝโโ๏ธ|๐ง๐พโโ๏ธ|๐ง๐ฟโโ๏ธ|๐ถโ๐ซ๏ธ|๐ต๏ธโโ๏ธ|๐ต๏ธโโ๏ธ|๐๏ธโโ๏ธ|๐๏ธโโ๏ธ|๐๏ธโโ๏ธ|๐๏ธโโ๏ธ|๐ณ๏ธโ๐|๐ณ๏ธโโง๏ธ|โน๐ปโโ๏ธ|โน๐ผโโ๏ธ|โน๐ฝโโ๏ธ|โน๐พโโ๏ธ|โน๐ฟโโ๏ธ|โน๐ปโโ๏ธ|โน๐ผโโ๏ธ|โน๐ฝโโ๏ธ|โน๐พโโ๏ธ|โน๐ฟโโ๏ธ|๐ฎโ๐จ|๐ตโ๐ซ|โค๏ธโ๐ฅ|โค๏ธโ๐ฉน|๐งโโ๏ธ|๐งโโ๏ธ|๐จโ๐ฆฐ|๐จโ๐ฆฑ|๐จโ๐ฆณ|๐จโ๐ฆฒ|๐ฉโ๐ฆฐ|๐งโ๐ฆฐ|๐ฉโ๐ฆฑ|๐งโ๐ฆฑ|๐ฉโ๐ฆณ|๐งโ๐ฆณ|๐ฉโ๐ฆฒ|๐งโ๐ฆฒ|๐ฑโโ๏ธ|๐ฑโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐
โโ๏ธ|๐
โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐คฆโโ๏ธ|๐คฆโโ๏ธ|๐คทโโ๏ธ|๐คทโโ๏ธ|๐งโโ๏ธ|๐จโโ๏ธ|๐ฉโโ๏ธ|๐งโ๐|๐จโ๐|๐ฉโ๐|๐งโ๐ซ|๐จโ๐ซ|๐ฉโ๐ซ|๐งโโ๏ธ|๐จโโ๏ธ|๐ฉโโ๏ธ|๐งโ๐พ|๐จโ๐พ|๐ฉโ๐พ|๐งโ๐ณ|๐จโ๐ณ|๐ฉโ๐ณ|๐งโ๐ง|๐จโ๐ง|๐ฉโ๐ง|๐งโ๐ญ|๐จโ๐ญ|๐ฉโ๐ญ|๐งโ๐ผ|๐จโ๐ผ|๐ฉโ๐ผ|๐งโ๐ฌ|๐จโ๐ฌ|๐ฉโ๐ฌ|๐งโ๐ป|๐จโ๐ป|๐ฉโ๐ป|๐งโ๐ค|๐จโ๐ค|๐ฉโ๐ค|๐งโ๐จ|๐จโ๐จ|๐ฉโ๐จ|๐งโโ๏ธ|๐จโโ๏ธ|๐ฉโโ๏ธ|๐งโ๐|๐จโ๐|๐ฉโ๐|๐งโ๐|๐จโ๐|๐ฉโ๐|๐ฎโโ๏ธ|๐ฎโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐ทโโ๏ธ|๐ทโโ๏ธ|๐ณโโ๏ธ|๐ณโโ๏ธ|๐คตโโ๏ธ|๐คตโโ๏ธ|๐ฐโโ๏ธ|๐ฐโโ๏ธ|๐ฉโ๐ผ|๐จโ๐ผ|๐งโ๐ผ|๐งโ๐|๐ฆธโโ๏ธ|๐ฆธโโ๏ธ|๐ฆนโโ๏ธ|๐ฆนโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐ถโโ๏ธ|๐ถโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโ๐ฆฏ|๐จโ๐ฆฏ|๐ฉโ๐ฆฏ|๐งโ๐ฆผ|๐จโ๐ฆผ|๐ฉโ๐ฆผ|๐งโ๐ฆฝ|๐จโ๐ฆฝ|๐ฉโ๐ฆฝ|๐โโ๏ธ|๐โโ๏ธ|๐ฏโโ๏ธ|๐ฏโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|๐ฃโโ๏ธ|๐ฃโโ๏ธ|๐โโ๏ธ|๐โโ๏ธ|โน๏ธโโ๏ธ|โน๏ธโโ๏ธ|๐ดโโ๏ธ|๐ดโโ๏ธ|๐ตโโ๏ธ|๐ตโโ๏ธ|๐คธโโ๏ธ|๐คธโโ๏ธ|๐คผโโ๏ธ|๐คผโโ๏ธ|๐คฝโโ๏ธ|๐คฝโโ๏ธ|๐คพโโ๏ธ|๐คพโโ๏ธ|๐คนโโ๏ธ|๐คนโโ๏ธ|๐งโโ๏ธ|๐งโโ๏ธ|๐จโ๐ฆ|๐จโ๐ง|๐ฉโ๐ฆ|๐ฉโ๐ง|๐โ๐ฆบ|๐ปโโ๏ธ|๐ดโโ ๏ธ|๐โโฌ|๐ฆ๐จ|๐ฆ๐ฉ|๐ฆ๐ช|๐ฆ๐ซ|๐ฆ๐ฌ|๐ฆ๐ฎ|๐ฆ๐ฑ|๐ฆ๐ฒ|๐ฆ๐ด|๐ฆ๐ถ|๐ฆ๐ท|๐ฆ๐ธ|๐ฆ๐น|๐ฆ๐บ|๐ฆ๐ผ|๐ฆ๐ฝ|๐ฆ๐ฟ|๐ง๐ฆ|๐ง๐ง|๐ง๐ฉ|๐ง๐ช|๐ง๐ซ|๐ง๐ฌ|๐ง๐ญ|๐ง๐ฎ|๐ง๐ฏ|๐ง๐ฑ|๐ง๐ฒ|๐ง๐ณ|๐ง๐ด|๐ง๐ถ|๐ง๐ท|๐ง๐ธ|๐ง๐น|๐ง๐ป|๐ง๐ผ|๐ง๐พ|๐ง๐ฟ|๐จ๐ฆ|๐จ๐จ|๐จ๐ฉ|๐จ๐ซ|๐จ๐ฌ|๐จ๐ญ|๐จ๐ฎ|๐จ๐ฐ|๐จ๐ฑ|๐จ๐ฒ|๐จ๐ณ|๐จ๐ด|๐จ๐ต|๐จ๐ท|๐จ๐บ|๐จ๐ป|๐จ๐ผ|๐จ๐ฝ|๐จ๐พ|๐จ๐ฟ|๐ฉ๐ช|๐ฉ๐ฌ|๐ฉ๐ฏ|๐ฉ๐ฐ|๐ฉ๐ฒ|๐ฉ๐ด|๐ฉ๐ฟ|๐ช๐ฆ|๐ช๐จ|๐ช๐ช|๐ช๐ฌ|๐ช๐ญ|๐ช๐ท|๐ช๐ธ|๐ช๐น|๐ช๐บ|๐ซ๐ฎ|๐ซ๐ฏ|๐ซ๐ฐ|๐ซ๐ฒ|๐ซ๐ด|๐ซ๐ท|๐ฌ๐ฆ|๐ฌ๐ง|๐ฌ๐ฉ|๐ฌ๐ช|๐ฌ๐ซ|๐ฌ๐ฌ|๐ฌ๐ญ|๐ฌ๐ฎ|๐ฌ๐ฑ|๐ฌ๐ฒ|๐ฌ๐ณ|๐ฌ๐ต|๐ฌ๐ถ|๐ฌ๐ท|๐ฌ๐ธ|๐ฌ๐น|๐ฌ๐บ|๐ฌ๐ผ|๐ฌ๐พ|๐ญ๐ฐ|๐ญ๐ฒ|๐ญ๐ณ|๐ญ๐ท|๐ญ๐น|๐ญ๐บ|๐ฎ๐จ|๐ฎ๐ฉ|๐ฎ๐ช|๐ฎ๐ฑ|๐ฎ๐ฒ|๐ฎ๐ณ|๐ฎ๐ด|๐ฎ๐ถ|๐ฎ๐ท|๐ฎ๐ธ|๐ฎ๐น|๐ฏ๐ช|๐ฏ๐ฒ|๐ฏ๐ด|๐ฏ๐ต|๐ฐ๐ช|๐ฐ๐ฌ|๐ฐ๐ญ|๐ฐ๐ฎ|๐ฐ๐ฒ|๐ฐ๐ณ|๐ฐ๐ต|๐ฐ๐ท|๐ฐ๐ผ|๐ฐ๐พ|๐ฐ๐ฟ|๐ฑ๐ฆ|๐ฑ๐ง|๐ฑ๐จ|๐ฑ๐ฎ|๐ฑ๐ฐ|๐ฑ๐ท|๐ฑ๐ธ|๐ฑ๐น|๐ฑ๐บ|๐ฑ๐ป|๐ฑ๐พ|๐ฒ๐ฆ|๐ฒ๐จ|๐ฒ๐ฉ|๐ฒ๐ช|๐ฒ๐ซ|๐ฒ๐ฌ|๐ฒ๐ญ|๐ฒ๐ฐ|๐ฒ๐ฑ|๐ฒ๐ฒ|๐ฒ๐ณ|๐ฒ๐ด|๐ฒ๐ต|๐ฒ๐ถ|๐ฒ๐ท|๐ฒ๐ธ|๐ฒ๐น|๐ฒ๐บ|๐ฒ๐ป|๐ฒ๐ผ|๐ฒ๐ฝ|๐ฒ๐พ|๐ฒ๐ฟ|๐ณ๐ฆ|๐ณ๐จ|๐ณ๐ช|๐ณ๐ซ|๐ณ๐ฌ|๐ณ๐ฎ|๐ณ๐ฑ|๐ณ๐ด|๐ณ๐ต|๐ณ๐ท|๐ณ๐บ|๐ณ๐ฟ|๐ด๐ฒ|๐ต๐ฆ|๐ต๐ช|๐ต๐ซ|๐ต๐ฌ|๐ต๐ญ|๐ต๐ฐ|๐ต๐ฑ|๐ต๐ฒ|๐ต๐ณ|๐ต๐ท|๐ต๐ธ|๐ต๐น|๐ต๐ผ|๐ต๐พ|๐ถ๐ฆ|๐ท๐ช|๐ท๐ด|๐ท๐ธ|๐ท๐บ|๐ท๐ผ|๐ธ๐ฆ|๐ธ๐ง|๐ธ๐จ|๐ธ๐ฉ|๐ธ๐ช|๐ธ๐ฌ|๐ธ๐ญ|๐ธ๐ฎ|๐ธ๐ฏ|๐ธ๐ฐ|๐ธ๐ฑ|๐ธ๐ฒ|๐ธ๐ณ|๐ธ๐ด|๐ธ๐ท|๐ธ๐ธ|๐ธ๐น|๐ธ๐ป|๐ธ๐ฝ|๐ธ๐พ|๐ธ๐ฟ|๐น๐ฆ|๐น๐จ|๐น๐ฉ|๐น๐ซ|๐น๐ฌ|๐น๐ญ|๐น๐ฏ|๐น๐ฐ|๐น๐ฑ|๐น๐ฒ|๐น๐ณ|๐น๐ด|๐น๐ท|๐น๐น|๐น๐ป|๐น๐ผ|๐น๐ฟ|๐บ๐ฆ|๐บ๐ฌ|๐บ๐ฒ|๐บ๐ณ|๐บ๐ธ|๐บ๐พ|๐บ๐ฟ|๐ป๐ฆ|๐ป๐จ|๐ป๐ช|๐ป๐ฌ|๐ป๐ฎ|๐ป๐ณ|๐ป๐บ|๐ผ๐ซ|๐ผ๐ธ|๐ฝ๐ฐ|๐พ๐ช|๐พ๐น|๐ฟ๐ฆ|๐ฟ๐ฒ|๐ฟ๐ผ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐ค๐ป|๐ค๐ผ|๐ค๐ฝ|๐ค๐พ|๐ค๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐คฒ๐ป|๐คฒ๐ผ|๐คฒ๐ฝ|๐คฒ๐พ|๐คฒ๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐
๐ป|๐
๐ผ|๐
๐ฝ|๐
๐พ|๐
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๐ป|๐
๐ผ|๐
๐ฝ|๐
๐พ|๐
๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐คฆ๐ป|๐คฆ๐ผ|๐คฆ๐ฝ|๐คฆ๐พ|๐คฆ๐ฟ|๐คท๐ป|๐คท๐ผ|๐คท๐ฝ|๐คท๐พ|๐คท๐ฟ|๐ฎ๐ป|๐ฎ๐ผ|๐ฎ๐ฝ|๐ฎ๐พ|๐ฎ๐ฟ|๐ต๐ป|๐ต๐ผ|๐ต๐ฝ|๐ต๐พ|๐ต๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ฅท๐ป|๐ฅท๐ผ|๐ฅท๐ฝ|๐ฅท๐พ|๐ฅท๐ฟ|๐ท๐ป|๐ท๐ผ|๐ท๐ฝ|๐ท๐พ|๐ท๐ฟ|๐คด๐ป|๐คด๐ผ|๐คด๐ฝ|๐คด๐พ|๐คด๐ฟ|๐ธ๐ป|๐ธ๐ผ|๐ธ๐ฝ|๐ธ๐พ|๐ธ๐ฟ|๐ณ๐ป|๐ณ๐ผ|๐ณ๐ฝ|๐ณ๐พ|๐ณ๐ฟ|๐ฒ๐ป|๐ฒ๐ผ|๐ฒ๐ฝ|๐ฒ๐พ|๐ฒ๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐คต๐ป|๐คต๐ผ|๐คต๐ฝ|๐คต๐พ|๐คต๐ฟ|๐ฐ๐ป|๐ฐ๐ผ|๐ฐ๐ฝ|๐ฐ๐พ|๐ฐ๐ฟ|๐คฐ๐ป|๐คฐ๐ผ|๐คฐ๐ฝ|๐คฐ๐พ|๐คฐ๐ฟ|๐คฑ๐ป|๐คฑ๐ผ|๐คฑ๐ฝ|๐คฑ๐พ|๐คฑ๐ฟ|๐ผ๐ป|๐ผ๐ผ|๐ผ๐ฝ|๐ผ๐พ|๐ผ๐ฟ|๐
๐ป|๐
๐ผ|๐
๐ฝ|๐
๐พ|๐
๐ฟ|๐คถ๐ป|๐คถ๐ผ|๐คถ๐ฝ|๐คถ๐พ|๐คถ๐ฟ|๐ฆธ๐ป|๐ฆธ๐ผ|๐ฆธ๐ฝ|๐ฆธ๐พ|๐ฆธ๐ฟ|๐ฆน๐ป|๐ฆน๐ผ|๐ฆน๐ฝ|๐ฆน๐พ|๐ฆน๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ถ๐ป|๐ถ๐ผ|๐ถ๐ฝ|๐ถ๐พ|๐ถ๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐บ๐ป|๐บ๐ผ|๐บ๐ฝ|๐บ๐พ|๐บ๐ฟ|๐ด๐ป|๐ด๐ผ|๐ด๐ฝ|๐ด๐พ|๐ด๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ฃ๐ป|๐ฃ๐ผ|๐ฃ๐ฝ|๐ฃ๐พ|๐ฃ๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ด๐ป|๐ด๐ผ|๐ด๐ฝ|๐ด๐พ|๐ด๐ฟ|๐ต๐ป|๐ต๐ผ|๐ต๐ฝ|๐ต๐พ|๐ต๐ฟ|๐คธ๐ป|๐คธ๐ผ|๐คธ๐ฝ|๐คธ๐พ|๐คธ๐ฟ|๐คฝ๐ป|๐คฝ๐ผ|๐คฝ๐ฝ|๐คฝ๐พ|๐คฝ๐ฟ|๐คพ๐ป|๐คพ๐ผ|๐คพ๐ฝ|๐คพ๐พ|๐คพ๐ฟ|๐คน๐ป|๐คน๐ผ|๐คน๐ฝ|๐คน๐พ|๐คน๐ฟ|๐ง๐ป|๐ง๐ผ|๐ง๐ฝ|๐ง๐พ|๐ง๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐ญ๐ป|๐ญ๐ผ|๐ญ๐ฝ|๐ญ๐พ|๐ญ๐ฟ|๐ซ๐ป|๐ซ๐ผ|๐ซ๐ฝ|๐ซ๐พ|๐ซ๐ฟ|๐ฌ๐ป|๐ฌ๐ผ|๐ฌ๐ฝ|๐ฌ๐พ|๐ฌ๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|๐๐ป|๐๐ผ|๐๐ฝ|๐๐พ|๐๐ฟ|#๏ธโฃ|0๏ธโฃ|1๏ธโฃ|2๏ธโฃ|3๏ธโฃ|4๏ธโฃ|5๏ธโฃ|6๏ธโฃ|7๏ธโฃ|8๏ธโฃ|9๏ธโฃ|โ๐ป|โ๐ผ|โ๐ฝ|โ๐พ|โ๐ฟ|โ๐ป|โ๐ผ|โ๐ฝ|โ๐พ|โ๐ฟ|โ๐ป|โ๐ผ|โ๐ฝ|โ๐พ|โ๐ฟ|โ๐ป|โ๐ผ|โ๐ฝ|โ๐พ|โ๐ฟ|โ๐ป|โ๐ผ|โ๐ฝ|โ๐พ|โ๐ฟ|โน๐ป|โน๐ผ|โน๐ฝ|โน๐พ|โน๐ฟ|๐|๐|๐|๐|๐|๐
|๐คฃ|๐|๐|๐|๐|๐|๐|๐ฅฐ|๐|๐คฉ|๐|๐|๐|๐|๐ฅฒ|๐|๐|๐|๐คช|๐|๐ค|๐ค|๐คญ|๐คซ|๐ค|๐ค|๐คจ|๐|๐|๐ถ|๐|๐|๐|๐ฌ|๐คฅ|๐|๐|๐ช|๐คค|๐ด|๐ท|๐ค|๐ค|๐คข|๐คฎ|๐คง|๐ฅต|๐ฅถ|๐ฅด|๐ต|๐คฏ|๐ค |๐ฅณ|๐ฅธ|๐|๐ค|๐ง|๐|๐|๐|๐ฎ|๐ฏ|๐ฒ|๐ณ|๐ฅบ|๐ฆ|๐ง|๐จ|๐ฐ|๐ฅ|๐ข|๐ญ|๐ฑ|๐|๐ฃ|๐|๐|๐ฉ|๐ซ|๐ฅฑ|๐ค|๐ก|๐ |๐คฌ|๐|๐ฟ|๐|๐ฉ|๐คก|๐น|๐บ|๐ป|๐ฝ|๐พ|๐ค|๐บ|๐ธ|๐น|๐ป|๐ผ|๐ฝ|๐|๐ฟ|๐พ|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐งก|๐|๐|๐|๐|๐ค|๐ค|๐ค|๐ฏ|๐ข|๐ฅ|๐ซ|๐ฆ|๐จ|๐ณ|๐ฃ|๐ฌ|๐จ|๐ฏ|๐ญ|๐ค|๐|๐ค|๐|๐|๐|๐ค|๐ค|๐ค|๐ค|๐ค|๐ค|๐|๐|๐|๐|๐|๐|๐|๐|๐ค|๐ค|๐|๐|๐|๐คฒ|๐ค|๐|๐
|๐คณ|๐ช|๐ฆพ|๐ฆฟ|๐ฆต|๐ฆถ|๐|๐ฆป|๐|๐ง |๐ซ|๐ซ|๐ฆท|๐ฆด|๐|๐|๐
|๐|๐ถ|๐ง|๐ฆ|๐ง|๐ง|๐ฑ|๐จ|๐ง|๐ฉ|๐ง|๐ด|๐ต|๐|๐|๐
|๐|๐|๐|๐ง|๐|๐คฆ|๐คท|๐ฎ|๐ต|๐|๐ฅท|๐ท|๐คด|๐ธ|๐ณ|๐ฒ|๐ง|๐คต|๐ฐ|๐คฐ|๐คฑ|๐ผ|๐
|๐คถ|๐ฆธ|๐ฆน|๐ง|๐ง|๐ง|๐ง|๐ง|๐ง|๐ง|๐|๐|๐ถ|๐ง|๐ง|๐|๐|๐บ|๐ด|๐ฏ|๐ง|๐ง|๐คบ|๐|๐|๐|๐|๐ฃ|๐|๐|๐ด|๐ต|๐คธ|๐คผ|๐คฝ|๐คพ|๐คน|๐ง|๐|๐|๐ญ|๐ซ|๐ฌ|๐|๐|๐ช|๐ฃ|๐ค|๐ฅ|๐ซ|๐ฃ|๐ฆฐ|๐ฆฑ|๐ฆณ|๐ฆฒ|๐ต|๐|๐ฆ|๐ฆง|๐ถ|๐|๐ฆฎ|๐ฉ|๐บ|๐ฆ|๐ฆ|๐ฑ|๐|๐ฆ|๐ฏ|๐
|๐|๐ด|๐|๐ฆ|๐ฆ|๐ฆ|๐ฆฌ|๐ฎ|๐|๐|๐|๐ท|๐|๐|๐ฝ|๐|๐|๐|๐ช|๐ซ|๐ฆ|๐ฆ|๐|๐ฆฃ|๐ฆ|๐ฆ|๐ญ|๐|๐|๐น|๐ฐ|๐|๐ฟ|๐ฆซ|๐ฆ|๐ฆ|๐ป|๐จ|๐ผ|๐ฆฅ|๐ฆฆ|๐ฆจ|๐ฆ|๐ฆก|๐พ|๐ฆ|๐|๐|๐ฃ|๐ค|๐ฅ|๐ฆ|๐ง|๐|๐ฆ
|๐ฆ|๐ฆข|๐ฆ|๐ฆค|๐ชถ|๐ฆฉ|๐ฆ|๐ฆ|๐ธ|๐|๐ข|๐ฆ|๐|๐ฒ|๐|๐ฆ|๐ฆ|๐ณ|๐|๐ฌ|๐ฆญ|๐|๐ |๐ก|๐ฆ|๐|๐|๐|๐ฆ|๐|๐|๐|๐ชฒ|๐|๐ฆ|๐ชณ|๐ท|๐ธ|๐ฆ|๐ฆ|๐ชฐ|๐ชฑ|๐ฆ |๐|๐ธ|๐ฎ|๐ต|๐น|๐ฅ|๐บ|๐ป|๐ผ|๐ท|๐ฑ|๐ชด|๐ฒ|๐ณ|๐ด|๐ต|๐พ|๐ฟ|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ฅญ|๐|๐|๐|๐|๐|๐|๐ซ|๐ฅ|๐
|๐ซ|๐ฅฅ|๐ฅ|๐|๐ฅ|๐ฅ|๐ฝ|๐ถ|๐ซ|๐ฅ|๐ฅฌ|๐ฅฆ|๐ง|๐ง
|๐|๐ฅ|๐ฐ|๐|๐ฅ|๐ฅ|๐ซ|๐ฅจ|๐ฅฏ|๐ฅ|๐ง|๐ง|๐|๐|๐ฅฉ|๐ฅ|๐|๐|๐|๐ญ|๐ฅช|๐ฎ|๐ฏ|๐ซ|๐ฅ|๐ง|๐ฅ|๐ณ|๐ฅ|๐ฒ|๐ซ|๐ฅฃ|๐ฅ|๐ฟ|๐ง|๐ง|๐ฅซ|๐ฑ|๐|๐|๐|๐|๐|๐|๐ |๐ข|๐ฃ|๐ค|๐ฅ|๐ฅฎ|๐ก|๐ฅ|๐ฅ |๐ฅก|๐ฆ|๐ฆ|๐ฆ|๐ฆ|๐ฆช|๐ฆ|๐ง|๐จ|๐ฉ|๐ช|๐|๐ฐ|๐ง|๐ฅง|๐ซ|๐ฌ|๐ญ|๐ฎ|๐ฏ|๐ผ|๐ฅ|๐ซ|๐ต|๐ถ|๐พ|๐ท|๐ธ|๐น|๐บ|๐ป|๐ฅ|๐ฅ|๐ฅค|๐ง|๐ง|๐ง|๐ง|๐ฅข|๐ฝ|๐ด|๐ฅ|๐ช|๐บ|๐|๐|๐|๐|๐บ|๐พ|๐งญ|๐|๐|๐ป|๐|๐|๐|๐|๐|๐|๐|๐|๐งฑ|๐ชจ|๐ชต|๐|๐|๐|๐ |๐ก|๐ข|๐ฃ|๐ค|๐ฅ|๐ฆ|๐จ|๐ฉ|๐ช|๐ซ|๐ฌ|๐ญ|๐ฏ|๐ฐ|๐|๐ผ|๐ฝ|๐|๐|๐|๐|๐|๐|๐|๐|๐
|๐|๐|๐|๐ |๐ก|๐ข|๐|๐ช|๐|๐|๐|๐
|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ป|๐|๐|๐|๐|๐|๐ต|๐ฆฝ|๐ฆผ|๐บ|๐ฒ|๐ด|๐น|๐ผ|๐|๐ฃ|๐ค|๐ข|๐จ|๐ฅ|๐ฆ|๐|๐ง|๐ถ|๐ค|๐ณ|๐ฅ|๐ข|๐ฉ|๐ซ|๐ฌ|๐ช|๐บ|๐|๐|๐ |๐ก|๐ฐ|๐|๐ธ|๐|๐งณ|๐ฐ|๐|๐ง|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ |๐|๐ก|๐|๐ข|๐|๐ฃ|๐|๐ค|๐|๐ฅ|๐|๐ฆ|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ก|๐|๐|๐ช|๐|๐ |๐|๐ค|๐ฅ|๐ฆ|๐ง|๐จ|๐ฉ|๐ช|๐ซ|๐ฌ|๐|๐|๐|๐ฅ|๐ง|๐|๐|๐|๐|๐|๐งจ|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐งง|๐|๐|๐|๐|๐ซ|๐|๐|๐
|๐ฅ|๐ฅ|๐ฅ|๐ฅ|๐|๐|๐|๐|๐พ|๐ฅ|๐ณ|๐|๐|๐|๐ฅ|๐|๐ธ|๐ฅ|๐ฅ|๐ฅ
|๐ฃ|๐คฟ|๐ฝ|๐ฟ|๐ท|๐ฅ|๐ฏ|๐ช|๐ช|๐ฑ|๐ฎ|๐ช|๐งฟ|๐ฎ|๐น|๐ฐ|๐ฒ|๐งฉ|๐งธ|๐ช
|๐ช|๐|๐|๐ด|๐ญ|๐ผ|๐จ|๐งต|๐ชก|๐งถ|๐ชข|๐|๐ถ|๐ฅฝ|๐ฅผ|๐ฆบ|๐|๐|๐|๐งฃ|๐งค|๐งฅ|๐งฆ|๐|๐|๐ฅป|๐ฉฑ|๐ฉฒ|๐ฉณ|๐|๐|๐|๐|๐|๐|๐|๐ฉด|๐|๐|๐ฅพ|๐ฅฟ|๐ |๐ก|๐ฉฐ|๐ข|๐|๐|๐ฉ|๐|๐งข|๐ช|๐ฟ|๐|๐|๐|๐|๐|๐|๐|๐ข|๐ฃ|๐ฏ|๐|๐|๐ผ|๐ต|๐ถ|๐|๐|๐|๐ค|๐ง|๐ป|๐ท|๐ช|๐ธ|๐น|๐บ|๐ป|๐ช|๐ฅ|๐ช|๐ฑ|๐ฒ|๐|๐|๐ |๐|๐|๐ป|๐ฅ|๐จ|๐ฑ|๐ฒ|๐ฝ|๐พ|๐ฟ|๐|๐งฎ|๐ฅ|๐|๐ฝ|๐ฌ|๐บ|๐ท|๐ธ|๐น|๐ผ|๐|๐|๐ฏ|๐ก|๐ฆ|๐ฎ|๐ช|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ฐ|๐|๐|๐|๐ท|๐ฐ|๐ช|๐ด|๐ต|๐ถ|๐ท|๐ธ|๐ณ|๐งพ|๐น|๐ง|๐จ|๐ฉ|๐ค|๐ฅ|๐ฆ|๐ซ|๐ช|๐ฌ|๐ญ|๐ฎ|๐ณ|๐|๐|๐|๐|๐|๐ผ|๐|๐|๐|๐
|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐จ|๐ช|๐ |๐ก|๐ซ|๐ช|๐น|๐ก|๐ช|๐ง|๐ช|๐ฉ|๐|๐ฆฏ|๐|๐ช|๐งฐ|๐งฒ|๐ช|๐งช|๐งซ|๐งฌ|๐ฌ|๐ญ|๐ก|๐|๐ฉธ|๐|๐ฉน|๐ฉบ|๐ช|๐|๐ช|๐ช|๐|๐|๐ช|๐ฝ|๐ช |๐ฟ|๐|๐ชค|๐ช|๐งด|๐งท|๐งน|๐งบ|๐งป|๐ชฃ|๐งผ|๐ชฅ|๐งฝ|๐งฏ|๐|๐ฌ|๐ชฆ|๐ฟ|๐ชง|๐ง|๐ฎ|๐ฐ|๐น|๐บ|๐ป|๐ผ|๐พ|๐|๐|๐|๐
|๐ธ|๐ซ|๐ณ|๐ญ|๐ฏ|๐ฑ|๐ท|๐ต|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐|๐ฏ|๐|๐|๐|๐ผ|๐ฝ|๐ฆ|๐
|๐|๐ถ|๐ณ|๐ด|๐ฑ|๐ฒ|๐ฑ|๐|๐ฐ|๐|๐ |๐ก|๐ข|๐ฃ|๐ค|๐
ฐ|๐|๐
ฑ|๐|๐|๐|๐|๐|๐|๐
พ|๐|๐
ฟ|๐|๐|๐|๐|๐|๐ท|๐ถ|๐ฏ|๐|๐น|๐|๐ฒ|๐|๐ธ|๐ด|๐ณ|๐บ|๐ต|๐ด|๐ |๐ก|๐ข|๐ต|๐ฃ|๐ค|๐ฅ|๐ง|๐จ|๐ฉ|๐ฆ|๐ช|๐ซ|๐ถ|๐ท|๐ธ|๐น|๐บ|๐ป|๐ |๐|๐ณ|๐ฒ|๐|๐ฉ|๐|๐ด|๐ณ|๐ป|๐ผ|๐ฝ|๐พ|๐ฟ|โบ|โน|โ |โฃ|โค|โ|โ|โ|โ|โ|โท|โน|โ|โ|โฐ|โช|โฉ|โฒ|โบ|โจ|โฝ|โ|โต|โด|โ|โ|โณ|โ|โฐ|โฑ|โฒ|โ|โญ|โ|โ
|โ|โ|โ|โฑ|โก|โ|โ|โ|โ|โจ|โฝ|โพ|โณ|โธ|โ |โฅ|โฆ|โฃ|โ|โ|โ|โจ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โฐ|โฑ|โฟ|โ |โ|โข|โฃ|โฌ|โ|โก|โ|โฌ|โ|โฌ
|โ|โ|โ|โฉ|โช|โคด|โคต|โ|โก|โธ|โฏ|โ|โฆ|โช|โฎ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โ|โถ|โฉ|โญ|โฏ|โ|โช|โฎ|โซ|โฌ|โธ|โน|โบ|โ|โ|โ|โง|โ|โ|โ|โ|โพ|โผ|โ|โ|โ|โ|โ|ใฐ|โ|โป|โ|โญ|โ
|โ|โ|โ|โ|โฐ|โฟ|ใฝ|โณ|โด|โ|ยฉ|ยฎ|โข|โน|โ|ใ|ใ|โซ|โช|โฌ|โฌ|โผ|โป|โพ|โฝ|โช|โซ)`
/*compile the pattern string into a regex*/
let emoRegex = new RegExp(emojiPattern, "g")
/*extracting the emojis*/
let emojis = [..."This ๐๐ฉโโ๏ธis the ๐งโโ๏ธtext๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ.".matchAll(emoRegex)];
console.log(emojis)
/*count of emojis*/
let emoCount = [..."This ๐๐ฉโโ๏ธis the ๐งโโ๏ธtext๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ.".matchAll(emoRegex)].length
console.log(emoCount)
/*strip emojis from text*/
let stripped = "This ๐๐ฉโโ๏ธis the ๐งโโ๏ธtext๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ.".replaceAll(emoRegex, "")
console.log(stripped)
/*use the pattern string to build a custom regex*/
let customRegex = new RegExp(".*"+emojiPattern+"{3}$") //match a string ending in 3 emojis
console.log(customRegex.test("yep three here ๐๐ฉโโ๏ธ๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ")) //true
console.log(customRegex.test("nope ๐ฅฃ๐ฉ๐ผโโค๏ธโ๐โ๐ฉ๐ฝ")) //false
You can use regular expression to detect it in input text:
/([\uE000-\uF8FF]|\uD83C[\uDF00-\uDFFF]|\uD83D[\uDC00-\uDDFF])/g
I wrote the following function: containsEmojis(input, includeBasic=true), which checks an input string for emojis according to the list of emojis defined in the Unicode specification version 13 (see https://unicode.org/Public/emoji/13.0/emoji-sequences.txt), and allows to ignore "basic" emojis that can be represented with only 3 bytes.
The snippet below defines the function and runs a few test cases:
/**
* iterates over the code points of an input string and returns true if an emoji is found.
*
* an emoji is found if the hex code for the character is 5 characters starting with "1F",
* or if #includeBasic is true, the character is 4 and starts with one of the prefixes of
* a basic emoji as defined in the Unicode specification version 13
* see https://unicode.org/Public/emoji/13.0/emoji-sequences.txt
*
* #input the string to check
* #includeBasic include also the basic emojis that only take 3 characters
*/
function containsEmojis(input, includeBasic) {
if (typeof includeBasic == "undefined")
includeBasic = true;
for (var c of input) {
var cHex = ("" + c).codePointAt(0).toString(16);
var lHex = cHex.length;
if (lHex > 3) {
var prefix = cHex.substring(0, 2);
if (lHex == 5 && prefix == "1f") {
return true;
}
if (includeBasic && lHex == 4) {
if (["20", "21", "23", "24", "25", "26", "27", "2B", "29", "30", "32"].indexOf(prefix) > -1)
return true;
}
}
}
return false;
}
// can be tested as follows:
var input;
input = "Hello World!";
console.log(input, containsEmojis(input));
input = "Hello ๐!";
console.log(input, containsEmojis(input));
console.log(input, containsEmojis(input, false));
// now try a basic emoji
input = "It sparkles โจ yay!";
console.log(input, containsEmojis(input));
// pass false for includeBasic
console.log(input, containsEmojis(input, false));
another solution I found that worked for me:
const example = '๐ฉโ๐ฉโ๐งโ๐ฆ๐ชฒ';
const regexpEmojiPresentation = /\p{Emoji_Presentation}/gu;
console.log(example.match(regexpEmojiPresentation));
check the emoji like bellow ways
function getEmojiChars(text) {
console.log(text.match(/\ud83c[\udf00-\udfff]|\ud83d[\udc00-\ude4f]|\ud83d[\ude80-\udeff]/g)) ;
}
you will get the array of emoji in the text
I have a string, msg. I want to check if the third character in the string exists or not.
I've tried if (msg.substring(3, 4) == undefined) { ... } but this doesn't work. What would the substring be equal to if it does not exist?
The third character of the string exists if the string has three characters, so the check you are looking for is
msg.length >= 3
As to your question about a non-existent substring, you get the empty string, which you can test in a console:
> "dog".substring(8, 13)
''
It's generally worth a mention that indexing and taking the length of characters is fraught with Unicode-related difficulties, because what JavaScript thinks of as a character is not really a character, but a UTF-16 character code. So watch out for:
> "๐ฌ๐ค".length
4
> "๐ฌ๐ค"[1]
'๏ฟฝ'
> "๐ฌ๐ค"[2]
'๏ฟฝ'
> [..."๐ฌ๐ค"][1]
'๐ค'
So if you are looking for better characters do:
[...msg].length >= 3
That will still not work with Unicode grapheme clusters, but it is at least better than taking the actual length of string, which is broken in JavaScript (unless you are dealing with what they call BMP characters).
From the documentation:
Any argument value that is less than 0 or greater than stringName.length is treated as if it were 0 and stringName.length, respectively.
This isn't super intuitive but your example will be an empty string "" (or if the first position has a character and you've added a larger gap, it will be a string that is shorter than the space between indexStart and indexEnd:
const str = 'hi';
console.log(str.substring(1, 4)); // is "i"
You could take a standard property accessor for getting a character or undefined.
const
string0 = '',
string1 = 'a',
string2 = 'ab',
string3 = 'abc',
string4 = 'abcd';
console.log(string0[2]); // undefined
console.log(string1[2]); // undefined
console.log(string2[2]); // undefined
console.log(string3[2]); // c
console.log(string4[2]); // c
If string doesn't contains enough letters substrings doesn't returns nothing. You can try use string as array which return undefined if index doesn't exist.
if ( msg[2] !== undefined ) {}
I have string patterns like name1|value1, name1|value1,name2|value2, name1| and name1|value1,. I have to have Regular expression to find the given pattern is true or false
Input and output would be
"name1|value1" -> true
"name1|value1,name2|value2" -> true
"name1|" -> false
"name1|value1," -> false
"name1|value1,name2" -> false
"name1|value1,name2|" -> false
Pretty simple: ^\w+\|\w+(,\w+\|\w+)*$
The first portion, ^\w+\|\w+ looks to make sure the string starts with at least 1 completed name|value pair.
Then the second portion, (,\w+\|\w+)* says that same pattern may repeat infinitely as long as there is a comma between the first pair and all subsequent pairs. (Although, the asterisk quantifies that the second portion of the pattern may not occur at all.)
Finally the $ says that the string must end matching this pattern. (I.e., this pattern cannot only match part of the string. It must match the entire string because of the ^ and $.)
To format this pattern for javascript, simply throw a forward slashes on both ends, so: /^\w+\|\w+(,\w+\|\w+)*$/ The pattern should not require any flags.
It is worth noting, if you need to match more complex names/values that are outside the character range of \w, then you should replace all \ws with [Some Character Set(s)].
If your have multiple pairs to check, you can apply your regex on splitted string elements with an every function:
isValidPairs = function(str) {
return str.split(',').every(function(elt) {
return /^\w+\|\w+$/.test(elt);
});
}
pairsArr = ["nam1|val1", "nam1|val1,name2|val2", "nam1|", "nam1|val1,", "nam1|val1,name", "nam1|val1,name|"];
pairsArr.forEach(function(str) {
console.log('%s: %s:', str, isValidPairs(str));
});
The use case is I want to compare a query string of characters to an array of words, and return the matches. A match is when a word contains all the characters in the query string, order doesn't matter, repeated characters are okay. Regex seems like it may be too powerful (a sledgehammer where only a hammer is needed). I've written a solution that compares the characters by looping through them and using indexOf, but it seems consistently slower. (http://jsperf.com/indexof-vs-regex-inside-a-loop/10) Is Regex the fastest option for this type of operation? Are there ways to make my alternate solution faster?
var query = "word",
words = ['word', 'wwoorrddss', 'words', 'argument', 'sass', 'sword', 'carp', 'drowns'],
reStoredMatches = [],
indexOfMatches = [];
function match(word, query) {
var len = word.length,
charMatches = [],
charMatch,
char;
while (len--) {
char = word[len];
charMatch = query.indexOf(char);
if (charMatch !== -1) {
charMatches.push(char);
}
}
return charMatches.length === query.length;
}
function linearIndexOf(words, query) {
var wordsLen = words.length,
wordMatch,
word;
while (wordsLen--) {
word = words[wordsLen];
wordMatch = match(word, query);
if (wordMatch) {
indexOfMatches.push(word);
}
}
}
function linearRegexStored(words, query) {
var wordsLen = words.length,
re = new RegExp('[' + query + ']', 'g'),
match,
word;
while (wordsLen--) {
word = words[wordsLen];
match = word.match(re);
if (match !== null) {
if (match.length >= query.length) {
reStoredMatches.push(word);
}
}
}
}
Note that your regex is wrong, that's most certainly why it goes so fast.
Right now, if your query is "word" (as in your example), the regex is going to be:
/[word]/g
This means look for one of the characters: 'w', 'o', 'r', or 'd'. If one matches, then match() returns true. Done. Definitively a lot faster than the most certainly more correct indexOf(). (i.e. in case of a simple match() call the 'g' flag is ignored since if any one thing matches, the function returns true.)
Also, you mention the idea/concept of any number of characters, I suppose as shown here:
'word', 'wwoorrddss'
The indexOf() will definitively not catch that properly if you really mean "any number" for each and every character. Because you should match an infinite number of cases. Something like this as a regex:
/w+o+r+d+s+/g
That you will certainly have a hard time to write the right code in plain JavaScript rather than use a regex. However, either way, that's going to be somewhat slow.
From the comment below, all the letters of the word are required, in order to do that, you have to have 3! tests (3 factorial) for a 3 letter word:
/(a.*b.*c)|(a.*c.*b)|(b.*a.*c)|(b.*c.*a)|(c.*a.*b)|(c.*b.*a)/
Obviously, a factorial is going to very quickly grow your number of possibilities and blow away your memory in a super long regex (although you can simplify if a word has the same letter multiple times, you do not have to test that letter more than once).
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
...
That's probably why your properly written test in plain JavaScript is much slower.
Also, in your case you should write the words nearly as done in Scrabble dictionaries: all letters once in alphabetical order (Scrabble keeps duplicates). So the word "word" would be "dorw". And as you shown in your example, the word "wwoorrddss" would be "dorsw". You can have some backend tool to generate your table of words (so you still write them as "word" and "words", and your tool massage those and convert them to "dorw" and "dorsw".) Then you can sort the letters of the words you are testing in alphabetical order and the result is that you do not need a silly factorial for the regex, you can simply do this:
/d.*o.*r.*w/
And that will match any word that includes the word "word" such as "password".
One easy way to sort the letters will be to split your word in an array of letters, and then sort the array. You may still get duplicates, it will depend on the sort capabilities. (I don't think that the default JavaScript sort will remove duplicates automatically.)
One more detail, if you test is supposed to be case insensitive, then you want to transform your strings to lowercase before running the test. So something like:
query = query.toLowerCase();
early on in your top function.
You are trying to speed up the algorithm "chars in word are a subset of the chars of query." You can short circuit this check and avoid some assignments (that are more readable but not strictly needed). Try the following version of match
function match(word, query) {
var len = word.length;
while (len--) {
if (query.indexOf(word[len]) === -1) { // found a missing char
return false;
}
}
return true; // couldn't find any missing chars
}
This gives a 4-5X improvement
Depending on the application you could try presorting words and presorting each word in words as another optimization.
The regexp match algorithm constructs a finite state automaton and makes its decisions on the current state and character read from left to right. This involves reading each character once and make a decision.
For static strings (to look a fixed string on a couple of text) you have better algorithms, like Knuth-Morris that allow you to go faster than one character at a time, but you must understand that this algorithm is not for matching regular expressions, just plain strings.
if you are interested in Knuth-Morris (there are several other algorithms) just have a round in wikipedia. http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm
A good thing you can do is to investigate if you regexp match routines do it with an DFA or a NDFA, as NDFAs occupy less memory and are easier to compute, but DFAs do it faster, but with some compilation penalties and more memory occupied.
Knuth-Morris algorithm also needs to compile the string into an automaton before working, so perhaps it doesn't apply to your problem if you are using it just to find one word in some string.