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Let's say I have a string like this:
...hello world.bye
But I want to remove the first three dots and replace .bye with !
So the output should be
hello world!
it should only match if both conditions apply (... at the beginning and .bye at the end)
And I'm trying to use js replace method. Could you please help? Thanks
First match the dots, capture and lazy-repeat any character until you get to .bye, and match the .bye. Then, you can replace with the first captured group, plus an exclamation mark:
const str = '...hello world.bye';
console.log(str.replace(/\.\.\.(.*)\.bye/, '$1!'));
The lazy-repeat is there to ensure you don't match too much, for example:
const str = `...hello world.bye
...Hello again! Goodbye.`;
console.log(str.replace(/\.\.\.(.*)\.bye/g, '$1!'));
You don't actually need a regex to do this. Although it's a bit inelegant, the following should work fine (obviously the function can be called whatever makes sense in the context of your application):
function manipulate(string) {
if (string.slice(0, 3) == "..." && string.slice(-4) == ".bye") {
return string.slice(4, -4) + "!";
}
return string;
}
(Apologies if I made any stupid errors with indexing there, but the basic idea should be obvious.)
This, to me at least, has the advantage of being easier to reason about than a regex. Of course if you need to deal with more complicated cases you may reach the point where a regex is best - but I personally wouldn't bother for a simple use-case like the one mentioned in the OP.
Your regex would be
const rx = /\.\.\.([\s\S]*?)\.bye/g
const out = '\n\nfoobar...hello world.bye\nfoobar...ok.bye\n...line\nbreak.bye\n'.replace(rx, `$1!`)
console.log(out)
In English, find three dots, anything eager in group, and ending with .bye.
The replacement uses the first match $1 and concats ! using a string template.
An arguably simpler solution:
const str = '...hello world.bye'
const newStr = /...(.+)\.bye/.exec(str)
const formatted = newStr ? newStr[1] + '!' : str
console.log(formatted)
If the string doesn't match the regex it will just return the string.
I want to extract only the first fontname out of a URL-string from the Google Webfont Directory. Here are some examples of possible strings and what part should be returned:
fonts.googleapis.com/css?family=Raleway // "Raleway"
fonts.googleapis.com/css?family=Caesar+Dressing // "Caesar Dressing"
fonts.googleapis.com/css?family=Raleway:300,400 // "Raleway"
fonts.googleapis.com/css?family=Raleway|Fondamento // "Raleway"
fonts.googleapis.com/css?family=Caesar+Dressing|Raleway:300,400|Fondamento // "Caesar Dressing"
So sometimes it's just one fontname, sometimes it has a weight indicated by a colon (:) and sometimes there are more fontnames divided by a pipe (|).
I have tried /family=(\S*)[:|]/ but it only matches the strings with :or |. I could do it like this, but it's not a nice solution:
var fontUrl = "fonts.googleapis.com/css?family=Caesar+Dressing|Raleway:300,400|Fondamento";
var fontName = /family=(\S*)/.exec(fontUrl)[1].replace(/\+/, " ");
if (fontName.indexOf(':') != -1){
fontName = fontName.split(':')[0];
}
if (fontName.indexOf('|') != -1){
fontName = fontName.split('|')[0];
}
console.log(fontName);
Is there a nice regex solution to this?
Instead of matching the character that (might) follow the string you want, match only the string you want except those characters:
/family=([^\s:|]*)/
Alternatively, you'd use a lookahead like this:
/family=(\S*?)(?=$|[:|])/
That should be better:
/family=([^:|]*)/
Of course for the + case, you'll have to replace it afterwards (or before maybe).
You can use (choose the i and m modifier in all case):
family=([a-z]+\+?[a-z]+)
or more simply
family=([a-z+]+)
or to avoid matching the + char:
family=([a-z]+)\+?([a-z]+)?
but it is an easyer way to use the second solution, and to replace the + chars with a space after.
try this:
/family\=(\S+?)[\:\|,]{0,2}\S*/ims
No regex is required in this case, unless you are good with regex's or test them thoroughly then you are likely to make mistakes.
var fontUrls = [];
fontUrls.push("fonts.googleapis.com/css?family=Raleway");
fontUrls.push("fonts.googleapis.com/css?family=Caesar+Dressing");
fontUrls.push("fonts.googleapis.com/css?family=Raleway:300,400");
fontUrls.push("fonts.googleapis.com/css?family=Raleway|Fondamento");
fontUrls.push("fonts.googleapis.com/css?family=Caesar+Dressing|Raleway:300,400|Fondamento");
function getFirstFont(url) {
return url.split("=")[1].split("|")[0].split(":")[0];
}
fontUrls.forEach(function (fontUrl) {
console.log(getFirstFont(fontUrl));
});
on jsfiddle
How do I remove emoji code using JavaScript? I thought I had taken care of it using the code below, but I still have characters like 🔴.
function removeInvalidChars() {
return this.replace(/[\uE000-\uF8FF]/g, '');
}
For me none of the answers completely removed all emojis so I had to do some work myself and this is what i got :
text.replace(/([\u2700-\u27BF]|[\uE000-\uF8FF]|\uD83C[\uDC00-\uDFFF]|\uD83D[\uDC00-\uDFFF]|[\u2011-\u26FF]|\uD83E[\uDD10-\uDDFF])/g, '');
Also, it should take into account that if one inserting the string later to the database, replacing with empty string could expose security issue. instead replace with the replacement character U+FFFD, see : http://www.unicode.org/reports/tr36/#Deletion_of_Noncharacters
The range you have selected is the Private Use Area, containing non-standard characters. Carriers used to encode emoji as different, inconsistent values inside this range.
More recently, the emoji have been given standardised 'unified' codepoints. Many of these are outside of the Basic Multilingual Plane, in the block U+1F300–U+1F5FF, including your example 🔴 U+1F534 Large Red Circle.
You could detect these characters with [\U0001F300-\U0001F5FF] in a regex engine that supported non-BMP characters, but JavaScript's RegExp is not such a beast. Unfortunately the JS string model is based on UTF-16 code units, so you'd have to work with the UTF-16 surrogates in a regexp:
return this.replace(/([\uE000-\uF8FF]|\uD83C[\uDF00-\uDFFF]|\uD83D[\uDC00-\uDDFF])/g, '')
However, note that there are other characters in the Basic Multilingual Plane that are used as emoji by phones but which long predate emoji. For example U+2665 is the traditional Heart Suit character ♥, but it may be rendered as an emoji graphic on some devices. It's up to you whether you treat this as emoji and try to remove it. See this list for more examples.
I solved it by using a regex with Unicode property escapes. I got it from this article, it's for Java but still very helpful - Remove Emojis from a Java String.
'Smile😀'.replace(/[^\p{L}\p{N}\p{P}\p{Z}^$\n]/gu, '');
It removes all symbols except:
\p{L} - all letters from any language
\p{N} - numbers
\p{P} - punctuation
\p{Z} - whitespace separators
^$\n - add any symbols you want to keep
This one should be more correct and it works, but for me it leaves some trash symbols in the string:
'Smile😀'.replace(/\p{Emoji}/gu, '');
Edit: added symbols from comments
I've found many suggestions around but the regex that have solved my problem is:
/(?:[\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
A short example
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, '');
}
Hope it can help you
Just an addition to #hababr answer.
If you need to get rid of complicated emojis, you have to remove also additional things like modifiers and etc:
'👨🏿🎤'.replace(/[\p{Emoji}\p{Emoji_Modifier}\p{Emoji_Component}\p{Emoji_Modifier_Base}\p{Emoji_Presentation}]/gu, '').charCodeAt(0)
update:
*#0-9 - are Emoji characters with a text representation by default, per the Unicode Standard.
so, my current solution is next:
'👨🏿🎤'.replace(/(?![*#0-9]+)[\p{Emoji}\p{Emoji_Modifier}\p{Emoji_Component}\p{Emoji_Modifier_Base}\p{Emoji_Presentation}]/gu, '').charCodeAt(0)
I know this post is a bit old, but I stumbled across this very problem at work and a colleague came up with an interesting idea. Basically instead of stripping emoji character only allow valid characters in. Consulting this ASCII table:
http://www.asciitable.com/
A function such as this could only keep legal characters (the range itself dependent on what you are after)
function (input) {
var result = '';
if (input.length == 0)
return input;
for (var indexOfInput = 0, lengthOfInput = input.length; indexOfInput < lengthOfInput; indexOfInput++) {
var charAtSpecificIndex = input[indexOfInput].charCodeAt(0);
if ((32 <= charAtSpecificIndex) && (charAtSpecificIndex <= 126)) {
result += input[indexOfInput];
}
}
return result;
};
This should preserve all numbers, letters and special characters of the Alphabet for a situation where you wish to preserve the English alphabet + number + special characters. Hope it helps someone :)
#bobince's solution didn't work for me. Either the Emojis stayed there or they were swapped by a different Emoji.
This solution did the trick for me:
var ranges = [
'\ud83c[\udf00-\udfff]', // U+1F300 to U+1F3FF
'\ud83d[\udc00-\ude4f]', // U+1F400 to U+1F64F
'\ud83d[\ude80-\udeff]' // U+1F680 to U+1F6FF
];
$('#mybtn').on('click', function() {
removeInvalidChars();
})
function removeInvalidChars() {
var str = $('#myinput').val();
str = str.replace(new RegExp(ranges.join('|'), 'g'), '');
$("#myinput").val(str);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="myinput"/>
<input type="submit" id="mybtn" value="clear"/>
Source
After searching and trying lots of unicode regex, I suggest you try this, it can cover all of emojis:
function removeEmoji(str) {
let strCopy = str;
const emojiKeycapRegex = /[\u0023-\u0039]\ufe0f?\u20e3/g;
const emojiRegex = /\p{Extended_Pictographic}/gu;
const emojiComponentRegex = /\p{Emoji_Component}/gu;
if (emojiKeycapRegex.test(strCopy)) {
strCopy = strCopy.replace(emojiKeycapRegex, '');
}
if (emojiRegex.test(strCopy)) {
strCopy = strCopy.replace(emojiRegex, '');
}
if (emojiComponentRegex.test(strCopy)) {
// eslint-disable-next-line no-restricted-syntax
for (const emoji of (strCopy.match(emojiComponentRegex) || [])) {
if (/[\d|*|#]/.test(emoji)) {
continue;
}
strCopy = strCopy.replace(emoji, '');
}
}
return strCopy;
}
let a = "1️⃣aa🤹♂️b#️⃣🔤✅❎23#!^*bb🤹🏾🤹♀️🚴🏻ccc";
console.log(removeEmoji(a))
Refrence: Unicode Emoij Document
None of the answers here worked for all the unicode characters I tested (specifically characters in the miscellaneous range such as ⛽ or ☯️).
Here is one that worked for me, (heavily) inspired from this SO PHP answer:
function _removeEmojis(str) {
return str.replace(/([#0-9]\u20E3)|[\xA9\xAE\u203C\u2047-\u2049\u2122\u2139\u3030\u303D\u3297\u3299][\uFE00-\uFEFF]?|[\u2190-\u21FF][\uFE00-\uFEFF]?|[\u2300-\u23FF][\uFE00-\uFEFF]?|[\u2460-\u24FF][\uFE00-\uFEFF]?|[\u25A0-\u25FF][\uFE00-\uFEFF]?|[\u2600-\u27BF][\uFE00-\uFEFF]?|[\u2900-\u297F][\uFE00-\uFEFF]?|[\u2B00-\u2BF0][\uFE00-\uFEFF]?|(?:\uD83C[\uDC00-\uDFFF]|\uD83D[\uDC00-\uDEFF])[\uFE00-\uFEFF]?/g, '');
}
(My use case is sorting in a data grid where emojis can come first in a string but users want the text ordered by the actual words.)
sandre89's answer is good but not perfect.
I spent some time on the subject and have a working solution.
var ranges = [
'[\u00A0-\u269f]',
'[\u26A0-\u329f]',
// The following characters could not be minified correctly
// if specifed with the ES6 syntax \u{1F400}
'[🀄-🧀]'
//'[\u{1F004}-\u{1F9C0}]'
];
$('#mybtn').on('click', function() {
removeInvalidChars();
});
function removeInvalidChars() {
var str = $('#myinput').val();
str = str.replace(new RegExp(ranges.join('|'), 'ug'), '');
$("#myinput").val(str);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="myinput" />
<input type="submit" id="mybtn" value="clear" />
Here is my CodePen
There are some points to note, though.
Unicode characters from U+1F000 up need a special notation, so you can use sandre89's way, or opt for the \u{1F000} ES6 notation, which may or may not work with your minificator. I succeeded pasting the emojis directly in the UTF-8 encoded script.
Don't forget the u flag in the regex, or your Javascript engine may throw an error.
Beware that things may not be working due to the file encoding, character set, or minificator. In my case nothing worked until I took the script off an .isml file (Demandware) and pasted it into a .js file.
You may gain some insight by referring to Wikipedia Emoji page and How many bytes does one Unicode character take?, and by tinkering with this Online Unicode converter, as I did.
var emoji =/([#0-9]\u20E3)|[\xA9\xAE\u203C\u2047-\u2049\u2122\u2139\u3030\u303D\u3297\u3299][\uFE00-\uFEFF]?|[\u2190-\u21FF][\uFE00-\uFEFF]?|[\u2300-\u23FF][\uFE00-\uFEFF]?|[\u2460-\u24FF][\uFE00-\uFEFF]?|[\u25A0-\u25FF][\uFE00-\uFEFF]?|[\u2600-\u27BF][\uFE00-\uFEFF]?|[\u2900-\u297F][\uFE00-\uFEFF]?|[\u2B00-\u2BF0][\uFE00-\uFEFF]?|(?:\uD83C[\uDC00-\uDFFF]|\uD83D[\uDC00-\uDEFF])[\uFE00-\uFEFF]?|[\u20E3]|[\u26A0-\u3000]|\uD83E[\udd00-\uddff]|[\u00A0-\u269F]/g;
str.replace(emoji, "");
i add this '\uD83E[\udd00-\uddff]'
these emojis were updated when 2018 june
if u want block emojis after other update then use this
str.replace(/[^0-9a-zA-Zㄱ-힣+×÷=%♤♡☆♧)(*&^/~##!-:;,?`_|<>{}¥£€$◇■□●○•°※¤《》¡¿₩\[\]\"\' \\]/g ,"");
u can block all emojis and u can only use eng, num, hangle, and some Characters
thx :)
You can use this function to replace emojis with nothing:
function msgAfterClearEmojis(msg)
{
var new_msg = msg.replace(/([#0-9]\u20E3)|[\xA9\xAE\u203C\u2047-\u2049\u2122\u2139\u3030\u303D\u3297\u3299][\uFE00-\uFEFF]?|[\u2190-\u21FF][\uFE00-\uFEFF]?|[\u2300-\u23FF][\uFE00-\uFEFF]?|[\u2460-\u24FF][\uFE00-\uFEFF]?|[\u25A0-\u25FF][\uFE00-\uFEFF]?|[\u2600-\u27BF][\uFE00-\uFEFF]?|[\u2900-\u297F][\uFE00-\uFEFF]?|[\u2B00-\u2BF0][\uFE00-\uFEFF]?|(?:\uD83C[\uDC00-\uDFFF]|\uD83D[\uDC00-\uDEFF])[\uFE00-\uFEFF]?|[\u20E3]|[\u26A0-\u3000]|\uD83E[\udd00-\uddff]|[\u00A0-\u269F]/g, '').trim();
return new_msg;
}
You can check here with emoji..
😊 , 😌 , 👽
function removeEmoji() {
var y = document.getElementById('textbox_id1');
y.value = y.value.replace(/([\u2700-\u27BF]|[\uE000-\uF8FF]|\uD83C[\uDC00-\uDFFF]|\uD83D[\uDC00-\uDFFF]|[\u2011-\u26FF]|\uD83E[\uDD10-\uDDFF])/g, '');
}
input {
padding: 5px;
}
<input type="text" id="textbox_id1" placeholder="Remove emoji..." oninput="removeEmoji()">
You can take more emojis from here: Emoji Keyboard Online
This is the iteration on #hababr's answer.
His answer removes lots of standard chars like $, +, < and so on.
This version keeps all of them (except for the \ backslash - dunno how to properly escape it).
"hey😁 hau💓 ahoy🏴☠️ !##$%^&*()-_=+±§;:'\|`~/?[]{},.<>".replace(/[^\p{L}\p{N}\p{P}\p{Z}{^$=+±\\'|`\\~<>}]/gu, "")
// "hey hau ahoy !##$%^&*()-_=+±§;:'|`~/?[]{},.<>"
I have this regex and it works for all emojis i found on this page
try this regex
<:[^:\s]+:\d+>|<a:[^:\s]+:\d+>|(\u00a9|\u00ae|[\u2000-\u3300]|\ud83c[\ud000-\udfff]|\ud83d[\ud000-\udfff]|\ud83e[\ud000-\udfff]|\ufe0f)
var emojiRegex = /\uD83C\uDFF4(?:\uDB40\uDC67\uDB40\uDC62(?:\uDB40\uDC65\uDB40\uDC6E\uDB40\uDC67|\uDB40\uDC77\uDB40\uDC6C\uDB40\uDC73|\uDB40\uDC73\uDB40\uDC63\uDB40\uDC74)\uDB40\uDC7F|\u200D\u2620\uFE0F)|\uD83D\uDC69\u200D\uD83D\uDC69\u200D(?:\uD83D\uDC66\u200D\uD83D\uDC66|\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67]))|\uD83D\uDC68(?:\u200D(?:\u2764\uFE0F\u200D(?:\uD83D\uDC8B\u200D)?\uD83D\uDC68|(?:\uD83D[\uDC68\uDC69])\u200D(?:\uD83D\uDC66\u200D\uD83D\uDC66|\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67]))|\uD83D\uDC66\u200D\uD83D\uDC66|\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67])|\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E[\uDDB0-\uDDB3])|(?:\uD83C[\uDFFB-\uDFFF])\u200D(?:\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E[\uDDB0-\uDDB3]))|\uD83D\uDC69\u200D(?:\u2764\uFE0F\u200D(?:\uD83D\uDC8B\u200D(?:\uD83D[\uDC68\uDC69])|\uD83D[\uDC68\uDC69])|\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E[\uDDB0-\uDDB3])|\uD83D\uDC69\u200D\uD83D\uDC66\u200D\uD83D\uDC66|(?:\uD83D\uDC41\uFE0F\u200D\uD83D\uDDE8|\uD83D\uDC69(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2695\u2696\u2708]|\uD83D\uDC68(?:(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2695\u2696\u2708]|\u200D[\u2695\u2696\u2708])|(?:(?:\u26F9|\uD83C[\uDFCB\uDFCC]|\uD83D\uDD75)\uFE0F|\uD83D\uDC6F|\uD83E[\uDD3C\uDDDE\uDDDF])\u200D[\u2640\u2642]|(?:\u26F9|\uD83C[\uDFCB\uDFCC]|\uD83D\uDD75)(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2640\u2642]|(?:\uD83C[\uDFC3\uDFC4\uDFCA]|\uD83D[\uDC6E\uDC71\uDC73\uDC77\uDC81\uDC82\uDC86\uDC87\uDE45-\uDE47\uDE4B\uDE4D\uDE4E\uDEA3\uDEB4-\uDEB6]|\uD83E[\uDD26\uDD37-\uDD39\uDD3D\uDD3E\uDDB8\uDDB9\uDDD6-\uDDDD])(?:(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2640\u2642]|\u200D[\u2640\u2642])|\uD83D\uDC69\u200D[\u2695\u2696\u2708])\uFE0F|\uD83D\uDC69\u200D\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67])|\uD83D\uDC69\u200D\uD83D\uDC69\u200D(?:\uD83D[\uDC66\uDC67])|\uD83D\uDC68(?:\u200D(?:(?:\uD83D[\uDC68\uDC69])\u200D(?:\uD83D[\uDC66\uDC67])|\uD83D[\uDC66\uDC67])|\uD83C[\uDFFB-\uDFFF])|\uD83C\uDFF3\uFE0F\u200D\uD83C\uDF08|\uD83D\uDC69\u200D\uD83D\uDC67|\uD83D\uDC69(?:\uD83C[\uDFFB-\uDFFF])\u200D(?:\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\uD83E[\uDDB0-\uDDB3])|\uD83D\uDC69\u200D\uD83D\uDC66|\uD83C\uDDF6\uD83C\uDDE6|\uD83C\uDDFD\uD83C\uDDF0|\uD83C\uDDF4\uD83C\uDDF2|\uD83D\uDC69(?:\uD83C[\uDFFB-\uDFFF])|\uD83C\uDDED(?:\uD83C[\uDDF0\uDDF2\uDDF3\uDDF7\uDDF9\uDDFA])|\uD83C\uDDEC(?:\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEE\uDDF1-\uDDF3\uDDF5-\uDDFA\uDDFC\uDDFE])|\uD83C\uDDEA(?:\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDED\uDDF7-\uDDFA])|\uD83C\uDDE8(?:\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDEE\uDDF0-\uDDF5\uDDF7\uDDFA-\uDDFF])|\uD83C\uDDF2(?:\uD83C[\uDDE6\uDDE8-\uDDED\uDDF0-\uDDFF])|\uD83C\uDDF3(?:\uD83C[\uDDE6\uDDE8\uDDEA-\uDDEC\uDDEE\uDDF1\uDDF4\uDDF5\uDDF7\uDDFA\uDDFF])|\uD83C\uDDFC(?:\uD83C[\uDDEB\uDDF8])|\uD83C\uDDFA(?:\uD83C[\uDDE6\uDDEC\uDDF2\uDDF3\uDDF8\uDDFE\uDDFF])|\uD83C\uDDF0(?:\uD83C[\uDDEA\uDDEC-\uDDEE\uDDF2\uDDF3\uDDF5\uDDF7\uDDFC\uDDFE\uDDFF])|\uD83C\uDDEF(?:\uD83C[\uDDEA\uDDF2\uDDF4\uDDF5])|\uD83C\uDDF8(?:\uD83C[\uDDE6-\uDDEA\uDDEC-\uDDF4\uDDF7-\uDDF9\uDDFB\uDDFD-\uDDFF])|\uD83C\uDDEE(?:\uD83C[\uDDE8-\uDDEA\uDDF1-\uDDF4\uDDF6-\uDDF9])|\uD83C\uDDFF(?:\uD83C[\uDDE6\uDDF2\uDDFC])|\uD83C\uDDEB(?:\uD83C[\uDDEE-\uDDF0\uDDF2\uDDF4\uDDF7])|\uD83C\uDDF5(?:\uD83C[\uDDE6\uDDEA-\uDDED\uDDF0-\uDDF3\uDDF7-\uDDF9\uDDFC\uDDFE])|\uD83C\uDDE9(?:\uD83C[\uDDEA\uDDEC\uDDEF\uDDF0\uDDF2\uDDF4\uDDFF])|\uD83C\uDDF9(?:\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDED\uDDEF-\uDDF4\uDDF7\uDDF9\uDDFB\uDDFC\uDDFF])|\uD83C\uDDE7(?:\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEF\uDDF1-\uDDF4\uDDF6-\uDDF9\uDDFB\uDDFC\uDDFE\uDDFF])|[#\*0-9]\uFE0F\u20E3|\uD83C\uDDF1(?:\uD83C[\uDDE6-\uDDE8\uDDEE\uDDF0\uDDF7-\uDDFB\uDDFE])|\uD83C\uDDE6(?:\uD83C[\uDDE8-\uDDEC\uDDEE\uDDF1\uDDF2\uDDF4\uDDF6-\uDDFA\uDDFC\uDDFD\uDDFF])|\uD83C\uDDF7(?:\uD83C[\uDDEA\uDDF4\uDDF8\uDDFA\uDDFC])|\uD83C\uDDFB(?:\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDEE\uDDF3\uDDFA])|\uD83C\uDDFE(?:\uD83C[\uDDEA\uDDF9])|(?:\uD83C[\uDFC3\uDFC4\uDFCA]|\uD83D[\uDC6E\uDC71\uDC73\uDC77\uDC81\uDC82\uDC86\uDC87\uDE45-\uDE47\uDE4B\uDE4D\uDE4E\uDEA3\uDEB4-\uDEB6]|\uD83E[\uDD26\uDD37-\uDD39\uDD3D\uDD3E\uDDB8\uDDB9\uDDD6-\uDDDD])(?:\uD83C[\uDFFB-\uDFFF])|(?:\u26F9|\uD83C[\uDFCB\uDFCC]|\uD83D\uDD75)(?:\uD83C[\uDFFB-\uDFFF])|(?:[\u261D\u270A-\u270D]|\uD83C[\uDF85\uDFC2\uDFC7]|\uD83D[\uDC42\uDC43\uDC46-\uDC50\uDC66\uDC67\uDC70\uDC72\uDC74-\uDC76\uDC78\uDC7C\uDC83\uDC85\uDCAA\uDD74\uDD7A\uDD90\uDD95\uDD96\uDE4C\uDE4F\uDEC0\uDECC]|\uD83E[\uDD18-\uDD1C\uDD1E\uDD1F\uDD30-\uDD36\uDDB5\uDDB6\uDDD1-\uDDD5])(?:\uD83C[\uDFFB-\uDFFF])|(?:[\u231A\u231B\u23E9-\u23EC\u23F0\u23F3\u25FD\u25FE\u2614\u2615\u2648-\u2653\u267F\u2693\u26A1\u26AA\u26AB\u26BD\u26BE\u26C4\u26C5\u26CE\u26D4\u26EA\u26F2\u26F3\u26F5\u26FA\u26FD\u2705\u270A\u270B\u2728\u274C\u274E\u2753-\u2755\u2757\u2795-\u2797\u27B0\u27BF\u2B1B\u2B1C\u2B50\u2B55]|\uD83C[\uDC04\uDCCF\uDD8E\uDD91-\uDD9A\uDDE6-\uDDFF\uDE01\uDE1A\uDE2F\uDE32-\uDE36\uDE38-\uDE3A\uDE50\uDE51\uDF00-\uDF20\uDF2D-\uDF35\uDF37-\uDF7C\uDF7E-\uDF93\uDFA0-\uDFCA\uDFCF-\uDFD3\uDFE0-\uDFF0\uDFF4\uDFF8-\uDFFF]|\uD83D[\uDC00-\uDC3E\uDC40\uDC42-\uDCFC\uDCFF-\uDD3D\uDD4B-\uDD4E\uDD50-\uDD67\uDD7A\uDD95\uDD96\uDDA4\uDDFB-\uDE4F\uDE80-\uDEC5\uDECC\uDED0-\uDED2\uDEEB\uDEEC\uDEF4-\uDEF9]|\uD83E[\uDD10-\uDD3A\uDD3C-\uDD3E\uDD40-\uDD45\uDD47-\uDD70\uDD73-\uDD76\uDD7A\uDD7C-\uDDA2\uDDB0-\uDDB9\uDDC0-\uDDC2\uDDD0-\uDDFF])|(?:[#\*0-9\xA9\xAE\u203C\u2049\u2122\u2139\u2194-\u2199\u21A9\u21AA\u231A\u231B\u2328\u23CF\u23E9-\u23F3\u23F8-\u23FA\u24C2\u25AA\u25AB\u25B6\u25C0\u25FB-\u25FE\u2600-\u2604\u260E\u2611\u2614\u2615\u2618\u261D\u2620\u2622\u2623\u2626\u262A\u262E\u262F\u2638-\u263A\u2640\u2642\u2648-\u2653\u265F\u2660\u2663\u2665\u2666\u2668\u267B\u267E\u267F\u2692-\u2697\u2699\u269B\u269C\u26A0\u26A1\u26AA\u26AB\u26B0\u26B1\u26BD\u26BE\u26C4\u26C5\u26C8\u26CE\u26CF\u26D1\u26D3\u26D4\u26E9\u26EA\u26F0-\u26F5\u26F7-\u26FA\u26FD\u2702\u2705\u2708-\u270D\u270F\u2712\u2714\u2716\u271D\u2721\u2728\u2733\u2734\u2744\u2747\u274C\u274E\u2753-\u2755\u2757\u2763\u2764\u2795-\u2797\u27A1\u27B0\u27BF\u2934\u2935\u2B05-\u2B07\u2B1B\u2B1C\u2B50\u2B55\u3030\u303D\u3297\u3299]|\uD83C[\uDC04\uDCCF\uDD70\uDD71\uDD7E\uDD7F\uDD8E\uDD91-\uDD9A\uDDE6-\uDDFF\uDE01\uDE02\uDE1A\uDE2F\uDE32-\uDE3A\uDE50\uDE51\uDF00-\uDF21\uDF24-\uDF93\uDF96\uDF97\uDF99-\uDF9B\uDF9E-\uDFF0\uDFF3-\uDFF5\uDFF7-\uDFFF]|\uD83D[\uDC00-\uDCFD\uDCFF-\uDD3D\uDD49-\uDD4E\uDD50-\uDD67\uDD6F\uDD70\uDD73-\uDD7A\uDD87\uDD8A-\uDD8D\uDD90\uDD95\uDD96\uDDA4\uDDA5\uDDA8\uDDB1\uDDB2\uDDBC\uDDC2-\uDDC4\uDDD1-\uDDD3\uDDDC-\uDDDE\uDDE1\uDDE3\uDDE8\uDDEF\uDDF3\uDDFA-\uDE4F\uDE80-\uDEC5\uDECB-\uDED2\uDEE0-\uDEE5\uDEE9\uDEEB\uDEEC\uDEF0\uDEF3-\uDEF9]|\uD83E[\uDD10-\uDD3A\uDD3C-\uDD3E\uDD40-\uDD45\uDD47-\uDD70\uDD73-\uDD76\uDD7A\uDD7C-\uDDA2\uDDB0-\uDDB9\uDDC0-\uDDC2\uDDD0-\uDDFF])\uFE0F|(?:[\u261D\u26F9\u270A-\u270D]|\uD83C[\uDF85\uDFC2-\uDFC4\uDFC7\uDFCA-\uDFCC]|\uD83D[\uDC42\uDC43\uDC46-\uDC50\uDC66-\uDC69\uDC6E\uDC70-\uDC78\uDC7C\uDC81-\uDC83\uDC85-\uDC87\uDCAA\uDD74\uDD75\uDD7A\uDD90\uDD95\uDD96\uDE45-\uDE47\uDE4B-\uDE4F\uDEA3\uDEB4-\uDEB6\uDEC0\uDECC]|\uD83E[\uDD18-\uDD1C\uDD1E\uDD1F\uDD26\uDD30-\uDD39\uDD3D\uDD3E\uDDB5\uDDB6\uDDB8\uDDB9\uDDD1-\uDDDD])/g;
console.log(text.replace(emojiRegex,'');
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script>
function isEmoji(str) {
var ranges = [
'[\uE000-\uF8FF]',
'\uD83C[\uDC00-\uDFFF]',
'\uD83D[\uDC00-\uDFFF]',
'[\u2011-\u26FF]',
'\uD83E[\uDD10-\uDDFF]'
];
if (str.match(ranges.join('|'))) {
return true;
} else {
return false;
}
}
$(document).ready(function(){
$('input').on('input',function(){
var $th = $(this);
console.log("Value of Input"+$th.val());
emojiInput= isEmoji($th.val());
if (emojiInput==true) {
$th.val("");
}
});
});
</script>
</head>
<body>
Enter your name: <input type="text">
</body>
</html>
There is a modern solution using categories
Modern browsers support Unicode property, which allows you to match emojis based on their belonging in the Emoji Unicode category. For example, you can use Unicode property escapes like \p{Emoji} or \P{Emoji} to match/no match emoji characters. Note that 0123456789#* and other characters are interpreted as emojis using the previous Unicode category. Therefore, a better way to do this is to use the {Extended_Pictographic} Unicode category that denotes all the characters typically understood as emojis instead of the {Emoji} category.
const withEmojis = /\p{Extended_Pictographic}/u
withEmojis.test('😀😀');
//true
withEmojis.test('ab');
//false
withEmojis.test('1');
//false
or with negation
const noEmojis = /\P{Extended_Pictographic}/u
noEmojis.test('😀');
//false
noEmojis.test('1212');
//false
You can use mathiasbynens/emoji-regex package to remove or replace emojis.
You can see the latest build's content to grab the regex by visiting following url:
http://unpkg.com/emoji-regex/index.js
In detail, this function first uses TextEncoder to convert content into a byte array with utf-8 encoding, then loops through this array, if it finds a byte whose first five bits are 11110 (i.e. 0xF0), it means this is an emoji start, then it replaces this byte and the next three bytes with 0x30 (i.e. number 0). Finally, it uses TextDecoder to convert the modified byte array back to a string, and uses replaceAll method to remove extra 0s.
function removeEmoji (content) {
let conByte = new TextEncoder("utf-8").encode(content);
for (let i = 0; i < conByte.length; i++) {
if ((conByte[i] & 0xF8) == 0xF0) {
for (let j = 0; j < 4; j++) {
conByte[i+j]=0x30;
}
i += 3;
}
}
content = new TextDecoder("utf-8").decode(conByte);
return content.replaceAll("0000", "");
}
I have tried to delete an item from a string divided with commas:
var str="this,is,unwanted,a,test";
if I do a simple str.replace('unwanted',''); I end up with 2 commas
if I do a more complex str.replace('unwanted','').replace(',,','');
It might work
But the problem comes when the str is like this:
var str="unwanted,this,is,a,test"; // or "...,unwanted"
However, I could do a 'if char at [0 or str.length] == comma', then remove it
But I really think this is not the way to go, it is absurd I need to do 2 replaces and 2 ifs to achieve what I want
I have heard that regex can do powerful stuff, but I simply can't understand it no matter how hard I try
Important Notes:
It should match after OR before (not both), or we will end with
"this,is,,a,test"
There are no spaces between commas
How about something less flaky than a regex for this sort of replacement?
str = str
.split(',')
.filter(function(token) { return token !== 'unwanted' })
.join(',');
jsFiddle.
However if you are convinced a regex is the best way...
str = str.replace(/(^|,)?unwanted(,|$)?/g, function(all, leading, trailing) {
return leading && trailing ? ',' : '';
});
(thanks Logan F. Smyth.)
jsFiddle.
Since Alex hasn't fixed this in his solution, I wanted to get a fully functional version up somewhere.
var unwanted = 'unwanted';
var regex = new RegExp('(^|,)' + unwanted + '(,|$)', 'g');
str = str.replace(regex, function(a, pre, suf) {
return pre && suf ? ',' : '';
});
The only thing to be careful of when dynamically building a regex, is that the 'unwanted' variable can't have anything in it that could be interpretted as a regex pattern.
There are way easier ways to parse this though, as Alex mentioned. Don't resort to regular expressions unless you have to.
I would like a RegExp that will remove all special characters from a string. I am trying something like this but it doesn’t work in IE7, though it works in Firefox.
var specialChars = "!##$^&%*()+=-[]\/{}|:<>?,.";
for (var i = 0; i < specialChars.length; i++) {
stringToReplace = stringToReplace.replace(new RegExp("\\" + specialChars[i], "gi"), "");
}
A detailed description of the RegExp would be helpful as well.
var desired = stringToReplace.replace(/[^\w\s]/gi, '')
As was mentioned in the comments it's easier to do this as a whitelist - replace the characters which aren't in your safelist.
The caret (^) character is the negation of the set [...], gi say global and case-insensitive (the latter is a bit redundant but I wanted to mention it) and the safelist in this example is digits, word characters, underscores (\w) and whitespace (\s).
Note that if you still want to exclude a set, including things like slashes and special characters you can do the following:
var outString = sourceString.replace(/[`~!##$%^&*()_|+\-=?;:'",.<>\{\}\[\]\\\/]/gi, '');
take special note that in order to also include the "minus" character, you need to escape it with a backslash like the latter group. if you don't it will also select 0-9 which is probably undesired.
Plain Javascript regex does not handle Unicode letters.
Do not use [^\w\s], this will remove letters with accents (like àèéìòù), not to mention to Cyrillic or Chinese, letters coming from such languages will be completed removed.
You really don't want remove these letters together with all the special characters. You have two chances:
Add in your regex all the special characters you don't want remove, for example: [^èéòàùì\w\s].
Have a look at xregexp.com. XRegExp adds base support for Unicode matching via the \p{...} syntax.
var str = "Їжак::: résd,$%& adùf"
var search = XRegExp('([^?<first>\\pL ]+)');
var res = XRegExp.replace(str, search, '',"all");
console.log(res); // returns "Їжак::: resd,adf"
console.log(str.replace(/[^\w\s]/gi, '') ); // returns " rsd adf"
console.log(str.replace(/[^\wèéòàùì\s]/gi, '') ); // returns " résd adùf"
<script src="https://cdnjs.cloudflare.com/ajax/libs/xregexp/3.1.1/xregexp-all.js"></script>
using \W or [a-z0-9] regex won't work for non english languages like chinese etc.,
It's better to use all special characters in regex and exclude them from given string
str.replace(/[~`!##$%^&*()+={}\[\];:\'\"<>.,\/\\\?-_]/g, '');
The first solution does not work for any UTF-8 alphabet. (It will cut text such as Їжак). I have managed to create a function which does not use RegExp and use good UTF-8 support in the JavaScript engine. The idea is simple if a symbol is equal in uppercase and lowercase it is a special character. The only exception is made for whitespace.
function removeSpecials(str) {
var lower = str.toLowerCase();
var upper = str.toUpperCase();
var res = "";
for(var i=0; i<lower.length; ++i) {
if(lower[i] != upper[i] || lower[i].trim() === '')
res += str[i];
}
return res;
}
Update: Please note, that this solution works only for languages where there are small and capital letters. In languages like Chinese, this won't work.
Update 2: I came to the original solution when I was working on a fuzzy search. If you also trying to remove special characters to implement search functionality, there is a better approach. Use any transliteration library which will produce you string only from Latin characters and then the simple Regexp will do all magic of removing special characters. (This will work for Chinese also and you also will receive side benefits by making Tromsø == Tromso).
I use RegexBuddy for debbuging my regexes it has almost all languages very usefull. Than copy/paste for the targeted language.
Terrific tool and not very expensive.
So I copy/pasted your regex and your issue is that [,] are special characters in regex, so you need to escape them. So the regex should be : /!##$^&%*()+=-[\x5B\x5D]\/{}|:<>?,./im
str.replace(/\s|[0-9_]|\W|[#$%^&*()]/g, "") I did sth like this.
But there is some people who did it much easier like str.replace(/\W_/g,"");
#Seagull anwser (https://stackoverflow.com/a/26482552/4556619)
looks good but you get undefined string in result when there are some special (turkish) characters. See example below.
let str="bənövşəyi 😟пурпурный İdÖĞ";
i slightly improve it and patch with undefined check.
function removeSpecials(str) {
let lower = str.toLowerCase();
let upper = str.toUpperCase();
let res = "",i=0,n=lower.length,t;
for(i; i<n; ++i) {
if(lower[i] !== upper[i] || lower[i].trim() === ''){
t=str[i];
if(t!==undefined){
res +=t;
}
}
}
return res;
}
text.replace(/[`~!##$%^*()_|+\-=?;:'",.<>\{\}\[\]\\\/]/gi, '');
why dont you do something like:
re = /^[a-z0-9 ]$/i;
var isValid = re.test(yourInput);
to check if your input contain any special char