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Possible to get 'regex source' from match?

I can get the source of a regex when it's defined separately. For example:
let r1 = new RegExp("el*");
console.log(r1.source);
// el*
Or:
let r2 = /el*/;
console.log(r2.source);
// el*
Is there a way to extract that if the regex isn't defined separately? For example, something along the lines of:
let m = "Hello".match(/el*/);
console.log(m.source?);

No,
quoting the documents of the match() function
Return value
An Array whose contents depend on the presence or absence of the
global (g) flag, or null if no matches are found.
So the return value is an array (you can test it by Array.isArray(m)// true)
However, the returned array has some extra information about the ocurred match (like groups, index and original input) but none of them include the original regex used to get the match
So there is no way to get that information from the match because its not returned by the matching function

The match result by itself cannot lead to the original regex, simply because different regexes can lead to the same result, even on the same string. Take for example the string "abcd" - all the following regexes: /abcd/, /a..d/ /a.*/ and many more, would match the string exactly the same way.
The only way you could retrive the original regex is if a reference to the regex was literally stored by the match() method inside the returned object. There is no reason to think that's the case, but you can implement your own match function that would do. Something like
function myMatch(str, regex) {
var match = str.match(regex);
if (match === null) {
match = [null];
}
match.source = regex;
return match;
}

Regexp group not excluding dots

Let's say I have the following string: div.classOneA.classOneB#idOne
Trying to write a regexp which extracts the classes (classOneA, classOneB) from it. I was able to do this but with Lookbehind assertion only.
It looks like this:
'div.classOneA.classOneB#idOne'.match(/(?<=\.)([^.#]+)/g)
> (2) ["classOneA", "classOneB"]
Now I would like to archive this without the lookbehind approach and do not really understand why my solution's not working.
'div.classOneA.classOneB#idOne'.match(/\.([^.#]+)/g)
> (2) [".classOneA", ".classOneB"]
Thought that the grouping will solve my problem but all matching item contains the dot as well.

There isn't a good way in Javascript to both match multiple times (/g option) and pick up capture groups (in the parens). Try this:
var input = "div.classOneA.classOneB#idOne";
var regex = /\.([^.#]+)/g;
var matches, output = [];
while (matches = regex.exec(input)) {
output.push(matches[1]);
}

This is because with g modifier you get all matching substrings but not its matching groups (that is as if (...) pairs worked just like (?:...) ones.
You see. Whithout g modifier:
> 'div.classOneA.classOneB#idOne'.match(/\.([^.#]+)/)
[ '.classOneA',
'classOneA',
index: 3,
input: 'div.classOneA.classOneB#idOne',
groups: undefined ]
With g modifier:
> 'div.classOneA.classOneB#idOne'.match(/\.([^.#]+)/g)
[ '.classOneA', '.classOneB' ]
In other words: you obtain all matches but only the whole match (0 item) per each.
There are many solutions:
Use LookBehind assertions as you pointed out yourself.
Fix each result later adding .map(x=>x.replace(/^\./, ""))
Or, if your input structure won't be much more complicated than the example you provide, simply use a cheaper approach:
> 'div.classOneA.classOneB#idOne'.replace(/#.*/, "").split(".").slice(1)
[ 'classOneA', 'classOneB' ]
Use .replace() + callback instead of .match() in order to be able to access capture groups of every match:
const str = 'div.classOneA.classOneB#idOne';
const matches = [];
str.replace(/\.([^.#]+)/g, (...args)=>matches.push(args[1]))
console.log(matches); // [ 'classOneA', 'classOneB' ]
I would recommend the third one (if there aren't other possible inputs that could eventually break it) because it is much more efficient (actual regular expressions are used only once to trim the '#idOne' part).

If you want to expand you regex. you can simply map on results and replace . with empty string
let op = 'div.classOneA.classOneB#idOne'.match(/\.([^.#]+)/g)
.map(e=> e.replace(/\./g,''))
console.log(op)

If you know you are searching for a text containing class, then you can use something like
'div.classOneA.classOneB#idOne'.match(/class[^.#]+/g)
If the only thing you know is that the text is preceded by a dot, then you must use lookbehind.

This regex will work without lookbehind assertion:
'div.classOneA.classOneB#idOne'.match(/\.[^\.#]+/g).map(item => item.substring(1));
Lookbehind assertion is not available in JavaScript recently.

I'm not an expert on using regex - particularly in Javascript - but after some research on MDN I've figured out why your attempt wasn't working, and how to fix.
The problem is that using .match with a regexp with the /g flag will ignore capturing groups. So instead you have to use the .exec method on the regexp object, using a loop to execute it multiple times to get all the results.
So the following code is what works, and can be adapted for similar cases. (Note the grp[1] - this is because the first element of the array returned by .exec is the entire match, the groups are the subsequent elements.)
var regExp = /\.([^.#]+)/g
var result = [];
var grp;
while ((grp = regExp.exec('div.classOneA.classOneB#idOne')) !== null) {
result.push(grp[1]);
}
console.log(result)

How can I inverse matched result of the pattern?

Here is my string:
Organization 2
info#something.org.au more#something.com market#gmail.com single#noidea.com
Organization 3
headmistress#money.com head#skull.com
Also this is my pattern:
/^.*?#[^ ]+|^.*$/gm
As you see in the demo, the pattern matches this:
Organization 2
info#something.org.au
Organization 3
headmistress#money.com
My question: How can I make it inverse? I mean I want to match this:
more#something.com market#gmail.com single#noidea.com
head#skull.com
How can I do that? Actually I can write a new (and completely different) pattern to grab expected result, but I want to know, Is "inverting the result of a pattern" possible?

No, I don't believe there is a way to directly inverse a Regular Expression but keeping it the same otherwise.
However, you could achieve something close to what you're after by using your existing RegExp to replace its matches with an empty string:
var everythingThatDidntMatchStr = str.replace(/^.*?#[^ ]+|^.*$/gm, '');

You can replace the matches from first RegExp by using Array.prototype.forEach() to replace matched RegExp with empty string using `String.ptototype.replace();
var re = str.match(/^.*?#[^ ]+|^.*$/gm);
var res = str;
re.forEach(val => res = res.replace(new RegExp(val), ""));

Whats wrong with this regex logic

I am trying to fetch the value after equal sign, its works but i am getting duplicated values , any idea whats wrong here?
// Regex for finding a word after "=" sign
var myregexpNew = /=(\S*)/g;
// Regex for finding a word before "=" sign
var mytype = /(\S*)=/g;
//Setting data from Grid Column
var strNew = "QCById=20";
var matchNew = myregexpNew.exec(strNew);
var newtype = mytype.exec(strNew);
alert(matchNew);
https://jsfiddle.net/6vjjv0hv/

exec returns an array, the first element is the global match, the following ones are the submatches, that's why you get ["=20", "20"] (using console.log here instead of alert would make it clearer what you get).
When looking for submatches and using exec, you're usually interested in the elements starting at index 1.
Regarding the whole parsing, it's obvious there are better solution, like using only one regex with two submatches, but it depends on the real goal.

You can try without using Regex like this:
var val = 'QCById=20';
var myString = val.substr(val.indexOf("=") + 1);
alert(myString);
Presently exec is returning you the matched value.

REGEXP.exec(SOMETHING) returns an array (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec).
The first item in the array is the full match and the rest matches the parenthesized substrings.

You do not get duplicated values, you just get an array of a matched value and the captured text #1.
See RegExp#exec() help:
If the match succeeds, the exec() method returns an array and updates properties of the regular expression object. The returned array has the matched text as the first item, and then one item for each capturing parenthesis that matched containing the text that was captured.
Just use the [1] index to get the captured text only.
var myregexpNew = /=(\S*)/g;
var strNew = "QCById=20";
var matchNew = myregexpNew.exec(strNew);
if (matchNew) {
console.log(matchNew[1]);
}
To get values on both sides of =, you can use /(\S*)=(\S*)/g regex:
var myregexpNew = /(\S*)=(\S*)/g;
var strNew = "QCById=20";
var matchNew = myregexpNew.exec(strNew);
if (matchNew) {
console.log(matchNew[1]);
console.log(matchNew[2]);
}
Also, you may want to add a check to see if the captured values are not undefined/empty since \S* may capture an empty string. OR use /(\S+)=(\S+)/g regex that requires at least one non-whitespace character to appear before and after the = sign.

Javascript Regexp - Match Characters after a certain phrase

I was wondering how to use a regexp to match a phrase that comes after a certain match. Like:
var phrase = "yesthisismyphrase=thisiswhatIwantmatched";
var match = /phrase=.*/;
That will match from the phrase= to the end of the string, but is it possible to get everything after the phrase= without having to modify a string?

You use capture groups (denoted by parenthesis).
When you execute the regex via match or exec function, the return an array consisting of the substrings captured by capture groups. You can then access what got captured via that array. E.g.:
var phrase = "yesthisismyphrase=thisiswhatIwantmatched";
var myRegexp = /phrase=(.*)/;
var match = myRegexp.exec(phrase);
alert(match[1]);
or
var arr = phrase.match(/phrase=(.*)/);
if (arr != null) { // Did it match?
alert(arr[1]);
}

phrase.match(/phrase=(.*)/)[1]
returns
"thisiswhatIwantmatched"
The brackets specify a so-called capture group. Contents of capture groups get put into the resulting array, starting from 1 (0 is the whole match).

It is not so hard, Just assume your context is :
const context = "https://example.com/pa/GIx89GdmkABJEAAA+AAAA";
And we wanna have the pattern after pa/, so use this code:
const pattern = context.match(/pa\/(.*)/)[1];
The first item include pa/, but for the grouping second item is without pa/, you can use each what you want.

Let try this, I hope it work
var p = /\b([\w|\W]+)\1+(\=)([\w|\W]+)\1+\b/;
console.log(p.test('case1 or AA=AA ilkjoi'));
console.log(p.test('case2 or AA=AB'));
console.log(p.test('case3 or 12=14'));

If you want to get value after the regex excluding the test phrase, use this:
/(?:phrase=)(.*)/
the result will be
0: "phrase=thisiswhatIwantmatched" //full match
1: "thisiswhatIwantmatched" //matching group