Develop Reference

Regex matches numbers in date but shouldn't

Why does my regex pattern match the date part of the string? It seems like I'm not accounting for the / (slash) correctly with [^\/] to avoid the pattern to match date strings?
const reg = new RegExp(
/(USD|\$|EUR|€|USDC|USDT)?\s?(\d+[^\/]|\d{1,3}(,\d{3})*)(\.\d+)?(k|K|m|M)?\b/,
"i"
);
const str = "02/22/2021 $50k";
console.log(reg.exec(str));
// result: ['02', undefined, '02', undefined, undefined, undefined, undefined, index: 0, input: '02/22/2021 $50k', groups: undefined]
// was expecting: [$50k,...]

You get those matches for the date part and the undefined ones, because you use a pattern with optional parts and alternations |
In your pattern there is this part (\d+[^\/]|\d{1,3}(,\d{3})*). That first part of the alternation \d+[^\/] matches 1+ digits followed by any char except a / (which can also match a digit) and the minimum amount of characters is 2. That part will match 20, 22 and 2021 in the date part.
If there is 1 digit, the second part of the alternation will match it.
If you want to match only numbers as well, you can assert not / to the left and the right, and make the whole part with the first alternatives like USD optional with the optional whitspace chars as well, to prevent matching that before only digits.
The last alternation can be shortened to a character class [km]? with a case insensitive flag.
See this page for the lookbehind support for Javascript.
(?:(?:USD|\$|EUR|€|USDC|USDT)\s?)?(?<!\/)\b(?:\d{1,3}(?:,\d{3})*(?:\.\d+)?|\d+)(?!\/)[KkMm]?\b
Regex demo
const reg = /(?:(?:USD|\$|EUR|€|USDC|USDT)\s?)?(?<!\/)\b(?:\d{1,3}(?:,\d{3})*(?:\.\d+)?|\d+)(?!\/)[KkMm]?\b/gi;
const str = "02/22/2021 $50k 1,213.3 11111111 $50,000 $50000"
const res = Array.from(str.matchAll(reg), m => m[0]);
console.log(res)
If the currency is not optional:
(?:USD|\$|EUR|€|USDC|USDT)\s?(?:\d{1,3}(?:,\d{3})*(?:\.\d+)?|\d+)[KkMm]?\b
Regex demo

I can't get your regex well. so i try to figure out what result you would expect. check this. in groups you have each part of your string.
const regex = /(\d{2})*\/?(\d{2})\/(\d{2,4})?\s*(USD|\$|EUR|€|USDC|USDT)?(\d*)(k|K|m|M)?\b/i
const regexNamed = /(?<day>\d{2})*\/?(?<month>\d{2})\/(?<year>\d{2,4})?\s*(?<currency>USD|\$|EUR|€|USDC|USDT)?(?<value>\d*)(?<unit>k|K|m|M)?\b/i
const str1 = '02/22/2021 $50k'
const str2 = '02/2021 €50m'
const m1 = str1.match(regex)
const m2 = str2.match(regexNamed)
console.log(m1)
console.log(m2.groups)
Blockquote

Multiple OR conditions for words in JavaScript regular expression

I trying to have a regular expression which is finding between two words but those words are not certain one.
2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞASınıfıE10/ENo303
This is my text. I'm trying to find the word between Soyadı and Sınıfı, in this case ERTANĞA, but the word Sınıfı also can be no, numara or any number. This is what I did.
soyad[ıi](.*)S[ıi]n[ıi]f[ıi]|no|numara|[0-9]
[ıi] is for Turkish character issue, don't mind that.

You can use something like below :
/.*Soyad(ı|i)|S(ı|i)n(ı|i)f(ı|i).*|no.*|numera.*|[0-9]/gmi
Here is the link I worked on : https://regex101.com/r/QXLjLF/1
In JS code:
const regex = /.*Soyad(ı|i)|S(ı|i)n(ı|i)f(ı|i).*|no.*|numera.*|[0-9]/gmi;
var str = `2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞASınıfıE10/ENo303`;
var newStr = str.replace(regex, '');
console.log(newStr);

You can use a single capture group to get the word ERTANĞA, keep the character class [ıi] instead of using an alternation for (ı|i) and group the alternatives at the end of the pattern using a non capture group (?:
soyad[ıi](.+?)(?:S[ıi]n[ıi]f[ıi]|n(?:o|umara)|[0-9])
soyad[ıi] Match soyadı or soyadi
(.+?) Capture group 1, match 1 or more chars as least as possible
(?: Non capture group
S[ıi]n[ıi]f[ıi] Match S and then ı or i etc..
| Or
n(?:o|umara) Match either no or numara
| Or
[0-9] Match a digit 0-9
) Close non capture group
Note that you don't need the /m flag as there are no anchors in the pattern.
Regex demo
const regex = /soyad[ıi](.+?)(?:S[ıi]n[ıi]f[ıi]|n(?:o|umara)|[0-9])/gi;
const str = "2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞASınıfıE10/ENo303\n";
console.log(Array.from(str.matchAll(regex), m => m[1]));

This might do it
const str = `2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞASınıfıE10/ENo303
2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞAnumaraE10/ENo303
2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞAnoE10/ENo303`
const re = /(?:Soyad(ı|i))(.*?)(?:S(ı|i)n(ı|i)f(ı|i)|no|numara)/gmi
console.log([...str.matchAll(re)].map(x => x[2]))
ES5
const str = `2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞASınıfıE10/ENo303
2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞAnumaraE10/ENo303
2015ÖĞLEYEMEKKARTI(2016-20.AdıMEVLÜTSoyadıERTANĞAnoE10/ENo303`
const re = /(?:Soyad(ı|i))(.*?)(?:S(ı|i)n(ı|i)f(ı|i)|no|numara)/gmi
const res = []
let match;
while ((match = re.exec(str)) !== null) res.push(match[2])
console.log(res)

Reg Exp for finding hashtag words

I have the following sentence as a test:
This is a test with #shouldshow and to see if there #show
#yes this#shouldnotshow what is going on here
I have figured out most of the Reg Exp I need. Here's what I have so far: /((?<=#)([A-Z]*))/gi
This matches every tag but also matches the shouldnotshow portion. I want to not match words that are prefixed by anything but # (excluding whitespace & \n).
So the only matched words I should get are: shouldshow show yes.
Note: after #show is a newline

You just need to see if the hash is prefixed with whitespace or starts the string
https://regex101.com/r/JDuGvr/1
/(\s|^)#(\w+)/gm
with positive lookbehind as OP used
https://regex101.com/r/06X3ZX/1
/(?<=(\s|^)#)(\w+)/gm;
use [a-zA-Z0-9] if you do not want an underscore
const re1 = /(\s|^)#(\w+)/gm;
const re2 = /(?<=(\s|^)#)(\w+)/gm;
const str = `This is a test with #shouldshow and to see if there #show
#yes this#shouldnotshow what is going on here`;
const res1 = [...str.matchAll(re1)].map(match => match[2]); // here the match is the third item
console.log(res1)
const res2 = [...str.matchAll(re2)].map(match => match[0]); // match is the first item
console.log(res2)

Another option could be using your pattern asserting a # on the left that does not have a non whitespace char before it using (?<!\S)# and get the match only without capture groups.
Match at least 1+ times a char A-Z to prevent matching an empty string.
(?<=(?<!\S)#)[A-Z]+
Regex demo
const regex = /(?<=(?<!\S)#)[A-Z]+/gi;
const str = `This is a test with #shouldshow and to see if there #show
#yes this#shouldnotshow what is going on her`;
console.log(str.match(regex));

Setting the end of the match

I have the following string:
[TITLE|prefix=a] [STORENAME|prefix=b|suffix=c] [DYNAMIC|limit=10|random=0|reverse=0]
And I would like to get the value of the prefix of TITLE, which is a.
I have tried it with (?<=TITLE|)(?<=prefix=).*?(?=]|\|) and that seems to work but that gives me also the prefix of STORENAME (b). So if [TITLE|prefix=a] will be missing in the string, I'll have the wrong value.
So I need to set the end of the match with ] that belongs to [TITLE. Please notice that this string is dynamic. So it could be [TITLE|suffix=x|prefix=y] as well.
const regex = "[TITLE|prefix=a] [STORENAME|prefix=b|suffix=c] [DYNAMIC|limit=10|random=0|reverse=0]".match(/(?<=TITLE|)(?<=prefix=).*?(?=]|\|)/);
console.log(regex);

You can use
(?<=TITLE(?:\|suffix=[^\]|]+)?\|prefix=)[^\]|]+
See the regex demo. Details:
(?<=TITLE(?:\|suffix=[^\]|]+)?\|prefix=) - a location in string immediately preceded with TITLE|prefix| or TITLE|suffix=...|prefix|
[^\]|]+ - one or more chars other than ] and |.
See JavaScript demo:
const texts = ['[TITLE|prefix=a] [STORENAME|prefix=b|suffix=c] [DYNAMIC|limit=10|random=0|reverse=0]', '[TITLE|suffix=s|prefix=a]'];
for (let s of texts) {
console.log(s, '=>', s.match(/(?<=TITLE(?:\|suffix=[^\]|]+)?\|prefix=)[^\]|]+/)[0]);
}

You could also use a capturing group
\[TITLE\|(?:[^|=\]]*=[^|=\]]*\|)*prefix=([^|=\]]*)[^\]]*]
Explanation
\[TITLE\| Match [TITLE|
(?:\w+=\w+\|)* Repeat 0+ occurrences wordchars = wordchars and |
prefix= Match literally
(\w+) Capture group 1, match 1+ word chars
[^\]]* Match any char except ]
] Match the closing ]
Regex demo
const regex = /\[TITLE\|(?:\w+=\w+\|)*prefix=(\w+)[^\]]*\]/g;
const str = `[TITLE|prefix=a] [STORENAME|prefix=b|suffix=c] [DYNAMIC|limit=10|random=0|reverse=0]
[TITLE|suffix=x|prefix=y]`;
let m;
while ((m = regex.exec(str)) !== null) {
console.log(m[1]);
}
Or with a negated character class instead of \w
\[TITLE\|(?:[^|=\]]*=[^|=\]]*\|)*prefix=([^|=\]]*)[^\]]*]
Regex demo

How to slice optional arguments in RegEx?

Actually i have the following RegExp expression:
/^(?:(?:\,([A-Za-z]{5}))?)+$/g
So the accepted input should be something like ,IGORA but even ,IGORA,GIANC,LOLLI is valid and i would be able to slice the string to 3 group in this case, in other the group number should be equals to the user input that pass the RegExp test.
i was trying to do something like this in JavaScript but it return only the last value
var str = ',GIANC,IGORA';
var arr = str.match(/^(?:(?:\,([A-Za-z]{5}))?)+$/).slice(1);
alert(arr);
So the output is 'IGORA' while i would it to be 'GIANC' 'IGORA'
Here is another example
/^([A-Z]{5})(?:(?:\,([A-Za-z]{2}))?)+$/g
test of regexp may have at least 5 chart string but it also can have other 5 chart string separated with a comma so from input
IGORA,CIAOA,POPOP
I would have an array of ["IGORA","CIAOA","POPOP"]

You can capture the words in a capturing surrounded by an optional preceding comma or an optional trailing comma.
You can test the regex here: ,?([A-Za-z]+),?
const pattern = /,?([A-Za-z]+),?/gm;
const str = `,IGORA,GIANC,LOLLI`;
let matches = [];
let match;
// Iterate until no match found
while ((m = pattern.exec(str))) {
// The first captured group is the match
matches.push(m[1]);
}
console.log(matches);

There are other ways to do this, but I found that one of the simple ways is by using the replace method, as it can replace all instances that match that regex.
For example:
var regex = /^(?:(?:\,([A-Za-z]{5}))?)+$/g;
var str = ',GIANC,IGORA';
var arr = [];
str.replace(regex, function(match) {
arr[arr.length] = match;
return match;
});
console.log(arr);
Also, in my code snippet you can see that there is an extra coma in each string, you can solve that by changing line 5 to arr[arr.length] = match.replace(/^,/, '').

Is this what you're looking for?
Explanation:
\b word boundary (starting or ending a word)
\w a word ([A-z])
{5} 5 characters of previous
So it matches all 5-character words but not NANANANA
var str = 'IGORA,CIAOA,POPOP,NANANANA';
var arr = str.match(/\b\w{5}\b/g);
console.log(arr); //['IGORA', 'CIAOA', 'POPOP']
If you only wish to select words separated by commas and nothing else, you can test for them like so:
(?<=,\s*|^) preceded by , with any number of trailing space, OR is the first word in list.
(?=,\s*|$) followed by , and any number of trailing spaces OR is last word in list.
In the following code, POPOP and MOMMA are rejected because they are not separated by a comma, and NANANANA fails because it is not 5 character.
var str = 'IGORA, CIAOA, POPOP MOMMA, NANANANA, MEOWI';
var arr = str.match(/(?<=,\s*|^)\b\w{5}\b(?=,\s*|$)/g);
console.log(arr); //['IGORA', 'CIAOA', 'MEOWI']
If you can't have any trailing spaces after the comma, just leave out the \s* from both (?<=,\s*|^) and (?=,\s*|$).