Regular expression for word boundaries but including emojis [duplicate] - javascript

This question already has an answer here:
Javascript Regular Expression for hashtags with emoji
(1 answer)
Closed 4 years ago.
I have a corpus of text that I'm parsing with a regular expression to find the most common words. Currently I'm using .match(/(?!'.*')\b\[\w'\]+\b/g). My problem is that \w does not match on non-alphanumeric characters and my emoji never get parsed. Specifically, I'm trying to make a regex that will identify words (including contractions) and emoji, separating on the word boundary.
As an example I'd like to be able to take "Hey there! 👋, let's go to the moon 🌝🚀" and get
Array( "Hey", "there", "👋", "let's", "go", "to", "the", "moon", "🌝", "🚀")

To work around it you may find spread operator helpful:
var str = "Hey there! 👋, let's go to the moon 🌝🚀";
var words = [], word = '';
[...str].forEach(function(char) {
// Test if current char is an English letter, a digit or '
if (/[a-z0-9']/i.test(char)) {
word += char;
}
// Test if current char is a non-whitespce char out of ASCII
else if (/(?=\S)[^\u0000-\u007f]/.test(char)) {
words.push(char)
} else if (word !== '') {
// Add a word to words array
words.push(word)
word = '';
}
})
console.log(words);

With guidance from commenters, I was able to get the following pattern to work:
\b[\w']+\b|\uD83D(?:[\uDC76\uDC66\uDC67](?:\uD83C[\uDFFB-\uDFFF])?|\uDC68(?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:\u2695\uFE0F?|\uD83C[\uDF93\uDFEB\uDF3E\uDF73\uDFED\uDFA4\uDFA8]|\u2696\uFE0F?|\uD83D[\uDD27\uDCBC\uDD2C\uDCBB\uDE80\uDE92]|\u2708\uFE0F?|\uD83E[\uDDB0-\uDDB3]))?)|\u200D(?:\u2695\uFE0F?|\uD83C[\uDF93\uDFEB\uDF3E\uDF73\uDFED\uDFA4\uDFA8]|\u2696\uFE0F?|\uD83D(?:\uDC69\u200D\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?)|\uDC68\u200D\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?)|\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?|[\uDD27\uDCBC\uDD2C\uDCBB\uDE80\uDE92])|\u2708\uFE0F?|\uD83E[\uDDB0-\uDDB3]|\u2764(?:\uFE0F\u200D\uD83D(?:\uDC8B\u200D\uD83D\uDC68|\uDC68)|\u200D\uD83D(?:\uDC8B\u200D\uD83D\uDC68|\uDC68)))))?|\uDC69(?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:\u2695\uFE0F?|\uD83C[\uDF93\uDFEB\uDF3E\uDF73\uDFED\uDFA4\uDFA8]|\u2696\uFE0F?|\uD83D[\uDD27\uDCBC\uDD2C\uDCBB\uDE80\uDE92]|\u2708\uFE0F?|\uD83E[\uDDB0-\uDDB3]))?)|\u200D(?:\u2695\uFE0F?|\uD83C[\uDF93\uDFEB\uDF3E\uDF73\uDFED\uDFA4\uDFA8]|\u2696\uFE0F?|\uD83D(?:\uDC69\u200D\uD83D(?:\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?)|\uDC66(?:\u200D\uD83D\uDC66)?|\uDC67(?:\u200D\uD83D[\uDC66\uDC67])?|[\uDD27\uDCBC\uDD2C\uDCBB\uDE80\uDE92])|\u2708\uFE0F?|\uD83E[\uDDB0-\uDDB3]|\u2764(?:\uFE0F\u200D\uD83D(?:\uDC8B\u200D\uD83D[\uDC68\uDC69]|[\uDC68\uDC69])|\u200D\uD83D(?:\uDC8B\u200D\uD83D[\uDC68\uDC69]|[\uDC68\uDC69])))))?|[\uDC74\uDC75](?:\uD83C[\uDFFB-\uDFFF])?|\uDC6E(?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|\uDD75(?:(?:\uFE0F(?:\u200D(?:[\u2642\u2640]\uFE0F?))?|\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|[\uDC82\uDC77](?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|\uDC78(?:\uD83C[\uDFFB-\uDFFF])?|\uDC73(?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|\uDC72(?:\uD83C[\uDFFB-\uDFFF])?|\uDC71(?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|[\uDC70\uDC7C](?:\uD83C[\uDFFB-\uDFFF])?|[\uDE4D\uDE4E\uDE45\uDE46\uDC81\uDE4B\uDE47\uDC86\uDC87\uDEB6](?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|[\uDC83\uDD7A](?:\uD83C[\uDFFB-\uDFFF])?|\uDC6F(?:\u200D(?:[\u2642\u2640]\uFE0F?))?|[\uDEC0\uDECC](?:\uD83C[\uDFFB-\uDFFF])?|\uDD74(?:(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F))?|\uDDE3\uFE0F?|[\uDEA3\uDEB4\uDEB5](?:(?:\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|[\uDCAA\uDC48\uDC49\uDC46\uDD95\uDC47\uDD96](?:\uD83C[\uDFFB-\uDFFF])?|\uDD90(?:(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F))?|[\uDC4C-\uDC4E\uDC4A\uDC4B\uDC4F\uDC50\uDE4C\uDE4F\uDC85\uDC42\uDC43](?:\uD83C[\uDFFB-\uDFFF])?|\uDC41(?:(?:\uFE0F(?:\u200D\uD83D\uDDE8\uFE0F?)?|\u200D\uD83D\uDDE8\uFE0F?))?|[\uDDE8\uDDEF\uDD73\uDD76\uDECD\uDC3F\uDD4A\uDD77\uDD78\uDDFA\uDEE3\uDEE4\uDEE2\uDEF3\uDEE5\uDEE9\uDEF0\uDECE\uDD70\uDD79\uDDBC\uDDA5\uDDA8\uDDB1\uDDB2\uDCFD\uDD6F\uDDDE\uDDF3\uDD8B\uDD8A\uDD8C\uDD8D\uDDC2\uDDD2\uDDD3\uDD87\uDDC3\uDDC4\uDDD1\uDDDD\uDEE0\uDDE1\uDEE1\uDDDC\uDECF\uDECB\uDD49]\uFE0F?|[\uDE00-\uDE06\uDE09-\uDE0B\uDE0E\uDE0D\uDE18\uDE17\uDE19\uDE1A\uDE42\uDE10\uDE11\uDE36\uDE44\uDE0F\uDE23\uDE25\uDE2E\uDE2F\uDE2A\uDE2B\uDE34\uDE0C\uDE1B-\uDE1D\uDE12-\uDE15\uDE43\uDE32\uDE41\uDE16\uDE1E\uDE1F\uDE24\uDE22\uDE2D\uDE26-\uDE29\uDE2C\uDE30\uDE31\uDE33\uDE35\uDE21\uDE20\uDE37\uDE07\uDE08\uDC7F\uDC79\uDC7A\uDC80\uDC7B\uDC7D\uDC7E\uDCA9\uDE3A\uDE38\uDE39\uDE3B-\uDE3D\uDE40\uDE3F\uDE3E\uDE48-\uDE4A\uDC64\uDC65\uDC6B-\uDC6D\uDC8F\uDC91\uDC6A\uDC63\uDC40\uDC45\uDC44\uDC8B\uDC98\uDC93-\uDC97\uDC99-\uDC9C\uDDA4\uDC9D-\uDC9F\uDC8C\uDCA4\uDCA2\uDCA3\uDCA5\uDCA6\uDCA8\uDCAB-\uDCAD\uDC53-\uDC62\uDC51\uDC52\uDCFF\uDC84\uDC8D\uDC8E\uDC35\uDC12\uDC36\uDC15\uDC29\uDC3A\uDC31\uDC08\uDC2F\uDC05\uDC06\uDC34\uDC0E\uDC2E\uDC02-\uDC04\uDC37\uDC16\uDC17\uDC3D\uDC0F\uDC11\uDC10\uDC2A\uDC2B\uDC18\uDC2D\uDC01\uDC00\uDC39\uDC30\uDC07\uDC3B\uDC28\uDC3C\uDC3E\uDC14\uDC13\uDC23-\uDC27\uDC38\uDC0A\uDC22\uDC0D\uDC32\uDC09\uDC33\uDC0B\uDC2C\uDC1F-\uDC21\uDC19\uDC1A\uDC0C\uDC1B-\uDC1E\uDC90\uDCAE\uDD2A\uDDFE\uDDFB\uDC92\uDDFC\uDDFD\uDD4C\uDD4D\uDD4B\uDC88\uDE82-\uDE8A\uDE9D\uDE9E\uDE8B-\uDE8E\uDE90-\uDE9C\uDEB2\uDEF4\uDEF9\uDEF5\uDE8F\uDEA8\uDEA5\uDEA6\uDED1\uDEA7\uDEF6\uDEA4\uDEA2\uDEEB\uDEEC\uDCBA\uDE81\uDE9F-\uDEA1\uDE80\uDEF8\uDD5B\uDD67\uDD50\uDD5C\uDD51\uDD5D\uDD52\uDD5E\uDD53\uDD5F\uDD54\uDD60\uDD55\uDD61\uDD56\uDD62\uDD57\uDD63\uDD58\uDD64\uDD59\uDD65\uDD5A\uDD66\uDD25\uDCA7\uDEF7\uDD2E\uDD07-\uDD0A\uDCE2\uDCE3\uDCEF\uDD14\uDD15\uDCFB\uDCF1\uDCF2\uDCDE-\uDCE0\uDD0B\uDD0C\uDCBB\uDCBD-\uDCC0\uDCFA\uDCF7-\uDCF9\uDCFC\uDD0D\uDD0E\uDCA1\uDD26\uDCD4-\uDCDA\uDCD3\uDCD2\uDCC3\uDCDC\uDCC4\uDCF0\uDCD1\uDD16\uDCB0\uDCB4-\uDCB8\uDCB3\uDCB9\uDCB1\uDCB2\uDCE7-\uDCE9\uDCE4-\uDCE6\uDCEB\uDCE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F2D-\uDF2F\uDF73\uDF72\uDF7F\uDF71\uDF58-\uDF5D\uDF60\uDF62-\uDF65\uDF61\uDF66-\uDF6A\uDF82\uDF70\uDF6B-\uDF6F\uDF7C\uDF75\uDF76\uDF7E\uDF77-\uDF7B\uDF74\uDFFA\uDF0D-\uDF10\uDF0B\uDFE0-\uDFE6\uDFE8-\uDFED\uDFEF\uDFF0\uDF01\uDF03-\uDF07\uDF09\uDF0C\uDFA0-\uDFA2\uDFAA\uDF11-\uDF20\uDF00\uDF08\uDF02\uDF0A\uDF83\uDF84\uDF86-\uDF8B\uDF8D-\uDF91\uDF80\uDF81\uDFAB\uDFC6\uDFC5\uDFC0\uDFD0\uDFC8\uDFC9\uDFBE\uDFB3\uDFCF\uDFD1-\uDFD3\uDFF8\uDFA3\uDFBD\uDFBF\uDFAF\uDFB1\uDFAE\uDFB0\uDFB2\uDCCF\uDC04\uDFB4\uDFAD\uDFA8\uDFBC\uDFB5\uDFB6\uDFA4\uDFA7\uDFB7-\uDFBB\uDFA5\uDFAC\uDFEE\uDFF9\uDFE7\uDFA6\uDD8E\uDD91-\uDD9A\uDE01\uDE36\uDE2F\uDE50\uDE39\uDE1A\uDE32\uDE51\uDE38\uDE34\uDE33\uDE3A\uDE35\uDFC1\uDF8C])|\u26F7\uFE0F?|\u26F9(?:(?:\uFE0F(?:\u200D(?:[\u2642\u2640]\uFE0F?))?|\uD83C(?:[\uDFFB-\uDFFF](?:\u200D(?:[\u2642\u2640]\uFE0F?))?)|\u200D(?:[\u2642\u2640]\uFE0F?)))?|[\u261D\u270C](?:(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F))?|[\u270B\u270A](?:\uD83C[\uDFFB-\uDFFF])?|\u270D(?:(?:\uD83C[\uDFFB-\uDFFF]|\uFE0F))?|[\u2764\u2763\u26D1\u2618\u26F0\u26E9\u2668\u26F4\u2708\u23F1\u23F2\u2600\u2601\u26C8\u2602\u26F1\u2744\u2603\u2604\u26F8\u2660\u2665\u2666\u2663\u260E\u2328\u2709\u270F\u2712\u2702\u26CF\u2692\u2694\u2699\u2696\u26D3\u2697\u26B0\u26B1\u26A0\u2622\u2623\u2B06\u2197\u27A1\u2198\u2B07\u2199\u2B05\u2196\u2195\u2194\u21A9\u21AA\u2934\u2935\u269B\u267E\u2721\u2638\u262F\u271D\u2626\u262A\u262E\u25B6\u23ED\u23EF\u25C0\u23EE\u23F8-\u23FA\u23CF\u2640\u2642\u2695\u267B\u269C\u2611\u2714\u2716\u303D\u2733\u2734\u2747\u203C\u2049\u3030\u00A9\u00AE\u2122]\uFE0F?|[\u0023\u002A\u0030-\u0039](?:\uFE0F\u20E3|\u20E3)|[\u2139\u24C2\u3297\u3299\u25AA\u25AB\u25FB\u25FC]\uFE0F?|[\u2615\u26EA\u26F2\u26FA\u26FD\u2693\u26F5\u231B\u23F3\u231A\u23F0\u2B50\u26C5\u2614\u26A1\u26C4\u2728\u26BD\u26BE\u26F3\u267F\u26D4\u2648-\u2653\u26CE\u23E9-\u23EC\u2B55\u2705\u274C\u274E\u2795-\u2797\u27B0\u27BF\u2753-\u2755\u2757\u25FD\u25FE\u2B1B\u2B1C\u26AA\u26AB]
which is basically word boundaries with all the emojis added in.
https://regex101.com/r/5rgYDx/3

Related

JavaScript replace regex with array element [duplicate]

This question already has an answer here:
Why this javascript regex doesn't work?
(1 answer)
Closed 3 years ago.
Trying to replace everything inside brackets [ ] with an element of an array. Example:
function replacingText(){
var names = ["Cole", "Kyle", "Chase"];
var sentance = 'This is [Cole].'
var regex = "\[(.*?)\]/gm";
console.log(sentance.replace(regex, names[1]));
}
So the output should be 'This is Kyle.' instead of 'This is [Cole].'
The only thing that needs fixed is the regex string needs to be
var regex = /\[(.*?)\]/gm;
The /gm on the end just means it wont stop at the first one it finds and the "m" stands for multi-line matching.
The javascript string replace can accept both strings and regular expressions as the first argument. See the examples presented here.
In your case you are passing the first as a string of a regular expression: "\[(.*?)\]"
Instead you should either match the exact string sentence.replace("[Cole]", names[1]) or, what you probably want, is to use the regular expression to match any name sentence.replace(/\[.+\]/g, names[1]) (note that the first argument does not contain any quotes)
The /g (global) is used to match all occurrences in the sentence. Otherwise only the first occurrence would be replaced.
Could you try this :
function replacingText() {
var names = ["Cole", "Kyle", "Chase"];
var sentance = "This is [Cole] [ahmed]";
var regex = /\[([0-9]|[aA-zZ])*\]/g;
console.log(sentance.replace(regex, names[1]));
}
I just tried it and it works as expected

Regexp Get Text Between parenttheis [duplicate]

This question already has answers here:
Regular Expression to get a string between parentheses in Javascript
(10 answers)
Closed 4 years ago.
I have string like this:
var str = "Hello (World) I'm Newbie";
how to get World from string above using RegExp?, I'm sorry I don't understand about regex.
Thank's
Rather than using a regex - use .split()...Note the escaped characters in the splits. The first split gives "World) I'm Newbie" and the second gives "World".
var str = "Hello (World) I'm Newbie";
var strContent = str.split('\(')[1].split('\)')[0];
console.log(strContent); // gives "World"
Assuming that there will be atleast one such word, you can do it using String#match. The following example matches the words between parentheses.
console.log(
"Hello (World) I'm Newbie"
.match(/\(\w+\)/g)
.map(match => match.slice(1, -1))
)
This might help you for your regex
\w match whole world
+ plus with another regex
[] starts group
^ except
(World) matching word
var str = "Hello (World) I'm Newbie";
var exactword=str.replace(/\w+[^(World)]/g,'')
var filtered = str.replace(/(World)/g,'')
alert(exactword)
alert(filtered)

RegExp() logical and and or result unexpected [duplicate]

This question already has answers here:
Regular Expressions: Is there an AND operator?
(14 answers)
Closed 7 years ago.
I have a simple filter function in my javascript, based on an input box.
function filter(selector, query) {
query = $.trim(query); //trim white space
query = query.replace(/ /gi, '|'); //add OR for regex
$(selector).each(function() {
($(this).text().search(new RegExp(query, "i")) < 0) ? (do something here)
So, if I have a table with a list of words, eg.
Alpha Centauri,
Beta Carbonate,
Charly Brown,
...
and I enter 'alpha cen' into my input box, the function searches for ('alpha' or 'cen') and returns the one record as desired.
However, if I replace the '|' with a '&' to search for ('alpha' and 'cen') I get no results. Even if I enter 'alpha centauri', I get no result at all.
Why?
While a | in a regex allows alternation, an & carries no special meaning. Your current code is trying to find a literal match for alpha&cen, which clearly doesn't match any of your data.

JavaScript reg ex apostrophe is new word boundry [duplicate]

This question already has answers here:
Convert string to Title Case with JavaScript
(68 answers)
Closed 7 years ago.
I'm trying to target the first letter of each word in a string. I'm working in JavaScript. I've found discussions for Python and PHP that don't address my problem specific to JavaScript.
str = "i'm a little tea pot"
regEx = /^.|\b[a-z]/g
I have a function that uppercases each letter the regEx finds. My result is the m after the apostrophe is capitalized. How do I avoid this?
After uppercase function:
I'M A Little Tea Pot"
Capitalize first letter of every word?
var string = prompt("Enter a string");
function capitalizeFirstWord(str) {
return str.split(/\s+/).map(function(word) {
return word[0].toUpperCase() + word.slice(1).toLowerCase();
}).join(' ');
}
document.getElementById("capitalized").innerHTML = capitalizeFirstWord(string);
<p id="capitalized"></p>
You split a string at a word boundary. As a parameter you pass in either an empty string or the whitespace character \s. This creates an array of all words in the sentence.
To get the first character, you can use word[0] or word.charAt(0). You then perform string concatenation with the rest of the string.
This gives you an entirely new array, which needs to be converted back to a string. Call the join method on this new array
You may use call-back function in str.replace.
var str = "i'm a little tea pot"
alert(str.replace(/(^|\s)([a-z])/g, function(match, group_1, group_2){return group_1 + group_2.toUpperCase()}))
And also if you want to convert the chars following the first char to lowercase then try this.
str.replace(/(^|\s)([a-z])(\S*)/g, function(x,y, z, a){return y + z.toUpperCase() + a.toLowerCase()})

Regex that searches for a starting character, anything in between, then an ending character? [closed]

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist
Closed 9 years ago.
Improve this question
So basically, I have a hugeee list of values that I copied from my site and would like them as just plain text. I need to create a regex for Javascript that removes the unwanted stuff
Here is what the values look like before hand:
<option value="111122223333">Some text (45)</option>
<option value="345835385390">Some text (10)</option>
<option value="456727235764">Some text (50)</option>
Here is what they need to look like afterwards:
Some text
Some text
Some text
Is this possible with 1 regex, or will many be needed?
My thinking is that a regex looks for < anything in between, and then > could be used on both the opening and closing option tags. Then the regex could be slightly modified to look for opening ( and then closing ) to remove the counter numbers to the left of the 'Some text' string.
Still learing regex, so it would be great if someone could also add a small explanation to their answer so that I could have some understanding of it. Thanks.
You can probably simplify this if you can use a library like jQuery. In that case you can use the text() method of a jQuery object to get the inner text, then run a simple regex to remove the '(xx)' part:
var vals = $('option').map(function() {
return $(this).text().replace(/\s*\(\d*\)$/, '');
});
// vals => ["some text", "some text", "some text"]
here's a fiddle: http://jsfiddle.net/jhummel/U46pH/
if you can't use a library I think you are looking for a regex like:
/<[^>]+>([^\(]+)\(\d*\)<[^>]+>/g
edit
You asked for a regex explanation, let's look at it part by part
/ = start the regex
< = find a '<' character
[^>]+ = find any character that is not a '>' one or more times
> = find a '>' character
( = start a group, anything in the parens will be saved for later
[^(]+ = find any character that is not a '(' one or more times - need to escape it with a backslash because the paren is a reserved character in regex
) = close the group
( = find a '(' char - need to escape it with a backslash again
\d* = find any numbers zero or more times
) = find a ')' character - escaped again
< = find a '<' character
[^>]+ = find any character that is note a '>' one or more times
> = find a '>' character
/ = end the regex
g = regex flag. Means find all the matches don't stop after the first match
If that's all you hope to accomplish then you can use something like:
(>)(.+)(<)
and then grab the second group out of the match.
EDIT: The parentheses are used to denote groups.

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