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Regex match hashtag with exceptions

I have the current expression:
/(?<![http://|https://|#])#([\d\w]+[^\d\s<]+[^\s<>]+)/g
However it's not compatible to run on Safari. I'm trying to handle the following cases:
#tag => match
#123 => no match
#32bit => match
##tag => no match
http://google.com/#/test => no match
tag##tag => no match
tag#tag => no match
<p>#tag</p> => match only #tag
#tag. => match only #tag
tag## => no match
tag# => no match
this is a match #tag => only #tag
I wonder how I can make a character before the match result in a negative match. E.g. # and /.
Is there any alternative to negative look behind that is compatible with Safari?
Thanks in advance.

You might use a negated character class and a capture group, and make sure that there are not only digits.
Note that \w also matches \d
(?:^|[^\w#/])(#(?!\d+\b)\w+)\b
Explanation
(?: Non capture group
^ Assert the start of the string
| Or
[^\w#/] Match a single non word char other than # or /
) Close non capture group
( Capture group 1
# Match literally
(?!\d+\b) Negative lookahead, assert not only digits to the right followed by a word boundary
\w+ Match 1+ word characters
) Close group 1
\b A word boundary to prevent a partial word match
Regex demo
let regex = /(?:^|[^\w#/])(#(?!\d+\b)\w+)\b/;
[
"#tag",
"#123",
"#32bit",
"##tag",
"http://google.com/#/test",
"tag##tag",
"tag#tag",
"<p>#tag</p>",
"#tag.",
"tag##",
"tag#"
].forEach(s => {
const m = s.match(regex)
if (m) {
console.log(`${s} --> ${m[1]}`)
}
})
Using the matches in a replacement:
let regex = /((?:^|[^\w#/]))(#(?!\d+\b)\w+)\b/;
[
"#tag",
"#123",
"#32bit",
"##tag",
"http://google.com/#/test",
"tag##tag",
"tag#tag",
"<p>#tag</p>",
"#tag.",
"tag##",
"tag#",
"this is a match #tag"
].forEach(s => {
const m = s.match(regex)
if (m) {
console.log(s.replace(regex, "$1<span>$2</span>"))
}
})

If you use the following pattern the second matching group contains what you want.
^(<\w*>)?(#\w+[a-zA-Z])

This satisfies your test cases. Not sure though whether you want this or not.
It does't work on #123 but I forgot it and I'm now lazy to add it as a screenshot.
/^(?:[^#]*[^#\w])?(#[\w]*[a-zA-Z][\w]*).*$/g
If you only want to allow <tags> before the "#", you can insted use #kendle's solution for the first non-capture group (before the actual group starting with #).
(?:<\w*>)?

You can also achieve this, without a capture group, with this regex:
/(?<![#\w])#{1}(?!\d+\b)\w+/g
const regex = /(?<![#\w])#{1}(?!\d+\b)\w+/g;
const stringToTest = [
"#tag",
"#123",
"#32bit",
"##tag",
"http://google.com/#/test",
"tag##tag",
"tag#tag",
"<p>#tag</p>",
"#tag.",
"tag##",
"tag#",
"this is a match #tag",
];
stringToTest.forEach(str => {
const match = str.match(regex);
if (match) {
console.log(`${str} -> ${match[0]}`);
} else {
console.log(`${str} -> ${match}`);
}
});
Good luck !

Javascript regex to find two characters between two delimitators

EDITED
I need to find two characters between '[' ']' and '/' '/' using Javascript.
I am using this regex:
([^.][/[string]]|\/string\/)|(\[(string))|(\/(string))| ((string)\])|((string)\/)
that gets two charactes but gets too one character.
The question is, how can I do to get just two characters?
Also I want to get exactly the two characters inside the string, I mean not just only the exact match.
Eg.
User input: dz
It must to find just exact matches that contains "dz", e.g. --> "dzone" but not "dazone". Currently I am getting matches with both strings, "dzone" and "dazone".
Demo: https://regex101.com/r/FEs6ib/1

You could optionally repeat any char except the delimiters between the delimiters them selves, and capture in a group what you want to keep.
If you want multiple matches for /dzone/dzone/ you could assert the last delimiter to the right instead of matching it.
The matches are in group 1 or group 2 where you can check for if they exist.
\/[^\/]*(dz)[^\/]*(?=\/)|\[[^\][]*(dz)[^\][]*(?=])
The pattern matches:
\/ Match /
[^\/]*(dz)[^\/]* Capture dz in group 1 between optional chars other than /
(?=\/) Positive lookahead, assert / to the right
| Or
\[ Match [
[^\][]*(dz)[^\][]* Capture dz in group 2 between optional chars other than [ and ]
-(?=]) Positive lookahead, assert ] to the right
Regex demo
This will match 1 occurrence of dz in the word. If you want to match the whole word, the capture group can be broadened to before and after the negated character class like:
\/([^\/]*dz[^\/]*)(?=\/)|\[([^\][]*dz[^\][]*)(?=])
Regex demo
const regex = /\/[^\/]*(dz)[^\/]*(?=\/)|\[[^\][]*(dz)[^\][]*(?=])/g;
[
"[dzone]",
"/dzone/",
"/dzone/dzone/",
"/testdztest/",
"[dazone]",
"/dazone/",
"dzone",
"dazone"
].forEach(s =>
console.log(
`${s} --> ${Array.from(s.matchAll(regex), m => m[2] ? m[2] : m[1])}`
)
);
If supported, you might also match all occurrences of dz between the delimiters using lookarounds with an infinite quantifier:
(?<=\/[^\/]*)dz(?=[^\/]*\/)|(?<=\[[^\][]*)dz(?=[^\][]*])
Regex demo
const regex = /(?<=\/[^\/]*)dz(?=[^\/]*\/)|(?<=\[[^\][]*)dz(?=[^\][]*])/g;
[
"[adzadzone]",
"[dzone]",
"/dzone/",
"/dzone/dzone/",
"/testdztest/",
"[dazone]",
"/dazone/",
"dzone",
"dazone"
].forEach(s => {
const m = s.match(regex);
if (m) {
console.log(`${s} --> ${s.match(regex)}`);
}
});

Regex add space in string if the word is longer than 4 characters and have numbers

I try to create a regex with 2 condition:
if word length more than 4 character
And if the word contains numbers
I need to add spaces
So like: iph12 return iph12, but iphone12 return iphone 12
I wrote regex
.replace(/\d+/gi, ' $& ').trim()
and this function return in anyway string like iphone 12. I tried to use function
.replace(/(?=[A-Z]+\d|\d+[A-Z])[A-Z\d]{,4}/i, ' $& ').trim()
but without second argument in {,4} it's not working. So is this possible?

You can use
text.replace(/\b([a-zA-Z]{4,})(\d+)\b/g, '$1 $2')
See the regex demo. Details:
\b - word boundary
([a-zA-Z]{4,}) - Group 1: four or more ASCII letters
(\d+) - Group 2: one or more digits
\b - word boundary
See the JavaScript demo:
const texts = ['iphone12', 'iph12'];
const regex = /\b([a-zA-Z]{4,})(\d+)\b/g;
for (const text of texts) {
console.log(text, '=>', text.replace(regex, '$1 $2'));
}
Output:
iphone12 => iphone 12
iph12 => iph12

Regex get string between ()

I have the text
var text = (hello) world this is (hi) text
I want to write a regex function so I can get
parseText(text) // returns ['hello', 'hi']
I tried this but not work:
'(hello) world this is (hi) text'.match('((.*?))')
Thanks for your help

you can try with:
/\([^\)]+\)/g
\(: escaped char
[^\)]+: one or more character(including symbols) until ) char.
\): escaped char
g flag: search all coincidences
const regex = /\([^\)]+\)/g;
const str = `(hello) world this is (hi) text`;
console.log(
str.match(regex) // this returns an string array
.map(i => i.slice(1, -1)) // remove first and last char
);
TIPS:
About point #2, you can change to [\)]* to take effect over zero
or more character.
If you need only string, you can use \w+ or \w*.
If you need only words you can use /\(\b\w+\b\)/g

You can find several options in this post.
Apart from using groups or postprocessing of the match results, you can use single regex match using lookahead / lookbehind:
var text = " (hello) world this is (hi) text"
var output = text.match(/(?<=\().*?(?=\))/g)
console.log(output)
output:
[ 'hello', 'hi' ]
Explanation:
(?<=...) ... positive lookbehind. The match is preceded be ..., but the ... is not included in the match
(?<=\() ... positive lookbehind for ( character
.* ... zero or more times of any character
.*? ... nongreedy version of .*
(?=...) ... positive lookahead, the match is followed by ... but the ... is not included in the match
(?=\)) ... positive lookahead for ) character
/.../g ... g is global flag, match finds all, not only the first, occurrence
do not forget to escape "special characters", e.g. parentheses

'(hello) world this is (hi) text'.match(/\([\w]*\)/g)
This returns [ "(hello)", "(hi)" ] and you can run another parse function to remove that extra parenthesis.
const text = '(hello) world this is (hi) text';
const list = text.match(/\([\w]*\)/g);
const parsed = list.map(item => item.replace(/\(|\)/g, ''));
console.log(parsed);

Replace match and ignore certain characters

I want to replace {r-group1} with "REPLACED" but leave the , where it is.
So, the string
var string = "{r-group1, }foo bar"
should output: "REPLACED, foo bar"
Using a negative lookahead, I tried adding a preceding (?![,]) group to leave the comma alone:
var replaced = string.replace(^(?:(?![,]){r-group1\})+$, 'REPLACED');
But it returns the same string. There are no matches to replace.
The same goes for a preceding comma:
var string = "foo bar{r-, group1}"
This should output: "foo bar, REPLACED"

You could do the replacement without a lookahead. You could match the curly braces and the content that comes before and after it except a comma using a negated character class [^,}]+ and capture the comma with optional whitespace chars in a capturing group.
In the replacement use the capturing groups $1REPLACED$2
Credits to #Nick for the updated pattern.
{r-(,?\s*)[^,}]+(,?\s*)}
Regex demo
const regex = /{r-(,?\s*)[^,}]+(,?\s*)}/g;
const str = `{r-group1, }foo bar`;
const subst = `$1REPLACED$2`;
const result = str.replace(regex, subst);
console.log(result);