Develop Reference

How to include a variable and exclude numbers[0-9] and letters[a-zA-Z] in RegExp?

I have a code that generates a random letter based on the word and I have tried to create a RegExp code to turn all the letters from the word to '_' except the randomly generated letter from the word.
const word = "Apple is tasty"
const randomCharacter = word[Math.floor(Math.random() * word.length)]
regex = new RegExp(/[^${randomCharacter}&\/\\#,+()$~%.'":;*?<>{}\s]/gi)
hint = word.replace(regex,'_')
I want to change all the letters to '_' except the randomly generated word. The above code for some reason does not work and shows the result: A___e __ ta_t_ and I'm not able to figure out what to do.
The final result I want is something like this: A____ __ _a___
Is there a way with regex to change all the alphabets and numbers '/[^a-zA-Z0-9]/g' to '_' except the randomly generated letter?
I'm listing all the expressions I want to include on my above code because I'm not able to figure out a way to do include and exclude at the same time using the variable with regex.

You can't do string interpolation inside of a RegExp literal (/.../). Meaning your placeholder ${randomCharacter} will not evaluate to its value in the template, but is instead interpreted literally as the string "${randomCharacter}".
If you want to use template literals, initialize your regex variable with a RegExp constructor instead, like:
const regex = new RegExp(`[^${randomCharacter}&\\/\\\#,+()$~%.'":;*?<>{}\\s]`, "gi");
See the MDN RegExp documentation for an explanation on the differences between the literal notation and constructor function, most notably:
The constructor of the regular expression object [...] results in runtime compilation of the regular expression. Use the constructor function when [...] you don't know the pattern and obtain it from another source, such as user input.

/(?:[^A\s])/
test it on regex101
just replace A in [^A\s] with you character that you want to ommit from replacement
demo:
const word = "Apple is tasty";
const randomCharacter = 'a';//word[Math.floor(Math.random() * word.length)];
regex = new RegExp('(?:[^' + randomCharacter + '\\s])', 'gi');
hint = word.replaceAll(regex, '_');
console.log(hint)

How to get next 3 characters after a substring if it exist inside a string?

I have this string that's. The &# substring is common but number after it changes in almost every object of my JSON data. So I want to detect if there is this substring, then get next three characters after it replace it with something else. How can I do it?

You can do it like this
var para = 'that's';
para = para.substr(para.indexOf('#')+1, 3);
syntax:
substr(start, length)
indexOf(searchvalue, [start])

Assuming you want to replace everything from &# until ; (if not, please update your question by specifying expected output):
You can use String.prototype.replace() with a regular expression:
var para = 'some string ' middle † end';
para = para.replace(/&#([\d]*);/g, 'replacement');
The g modifier is important to replace all occurences in the string.
With the RegEx used, you can include the found number (between &# and ;) in the replacement string by using $1.

you can define and use a utility function to replace HTML entities like that:
function decode(text, replaceWith = '') {
return text.replace(/&#(\d+);/g, replaceWith)
}

Extracting a complicated part of the string with plain Javascript

I have a following string:
Text
I want to extract from this string, with the use of JavaScript 'pl' or 'pl_company_com'
There are a few variables:
jan_kowalski is a name and surname it can change, and sometimes even have 3 elements
the country code (in this example 'pl') will change to other en / de / fr (this is that part of the string i want to get)
the rest of the string remains the same for every case (beginning + everything after starting with _company_com ...
Ps. I tried to do it with split, but my knowledge of JS is very basic and I cant get what i want, plase help

An alternative to Randy Casburn's solution using regex
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_(.*_company_com)')[1];
console.log(out);
Or if you want to just get that string with those country codes you specified
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_((en|de|fr|pl)_company_com)')[1];
console.log(out);
let out = new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx').href.match('.*_((en|de|fr|pl)_company_com)')[1];
console.log(out);
A proof of concept that this solution also works for other combinations
let urls = [
new URL('https://my.domain.com/personal/jan_kowalski_pl_company_com/Documents/Forms/All.aspx'),
new URL('https://my.domain.com/personal/firstname_middlename_lastname_pl_company_com/Documents/Forms/All.aspx')
]
urls.forEach(url => console.log(url.href.match('.*_(en|de|fr|pl).*')[1]))

I have been very successful before with this kind of problems with regular expressions:
var string = 'Text';
var regExp = /([\w]{2})_company_com/;
find = string.match(regExp);
console.log(find); // array with found matches
console.log(find[1]); // first group of regexp = country code
First you got your given string. Second you have a regular expression, which is marked with two slashes at the beginning and at the end. A regular expression is mostly used for string searches (you can even replace complicated text in all major editors with it, which can be VERY useful).
In this case here it matches exactly two word characters [\w]{2} followed directly by _company_com (\w indicates a word character, the [] group all wanted character types, here only word characters, and the {}indicate the number of characters to be found). Now to find the wanted part string.match(regExp) has to be called to get all captured findings. It returns an array with the whole captured string followed by all capture groups within the regExp (which are denoted by ()). So in this case you get the country code with find[1], which is the first and only capture group of the regular expression.

JavaScript String test with array of RegEx

I have some doubts regarding RegEx in JavaScript as I am not good in RegEx.
I have a String and I want to compare it against some array of RegEx expressions.
First I tried for one RegEx and it's not working. I want to fix that also.
function check(str){
var regEx = new RegEx("(users)\/[\w|\W]*");
var result = regEx.test(str);
if(result){
//do something
}
}
It is not working properly.
If I pass users, it doesn't match. If I pass users/ or users/somestring, it is matching.
If I change the RegEx to (usersGroupList)[/\w|\W]*, then it is matching for any string that contains the string users
fdgdsfgguserslist/data
I want to match like if string is either users or it should contain users/something or users/
And also I want the string to compare it with similar regex array.
I want to compare the string str with users, users/something, list, list/something, anothermatch, anothermatch/something. If if it matches any of these expression i want to do something.
How can I do that?
Thanks

Then, you'll have to make the last group optional. You do that by capturing the /something part in a group and following it with ? which makes the previous token, here the captured group, optional.
var regEx = new RegExp("(users)(\/[\w|\W]*)?");

What about making:
the last group optional
starting from beginning of the string
Like this:
var regEx = new RegExp("^(users)(\/[\w|\W]*)?");
Same applies for all the others cases, e.g. for list:
var regEx = new RegExp("^(list)(\/[\w|\W]*)?");
All in One Approach
var regEx = new RegExp("^(users|list|anothermatch)(\/[\w|\W]*)?");
Even More Generic
var keyw = ["users", "list", "anothermatch"];
var keyws = keyw.join("|");
var regEx = new RegExp("^("+keyws+")(\/[\w|\W]*)?");

You haven't made the / optional. Try this instead
(users)\/?[\w|\W]*

Using regex to split double hyphen but not single hyphen

I have an html element id that looks like this:
dp__1-2--1-3
I'm trying to use the JavaScript split() function to lop off and return the final '1-3'
My regex skills are poor but a bit of searching around got me to this point:
var myId = "dp__1-2--1-3";
var myIdPostFix = myId.split(/[\-\-]+/).pop();
Unfortunately that returns me only the '3'.
So my question is how do I split double hyphens but NOT single hyphens?

It's the brackets in the regular expression that keeps it from working. A set will match one of any of the characers in it, so [\-\-] is the same as [\-], i.e. matching a single hyphen.
Just remove the brackets:
var myIdPostFix = myId.split(/--/).pop();
or just use the string '--' instead of a regular expression:
var myIdPostFix = myId.split('--').pop();

split accepts a regular expression or a string as the first argument.
You were very close. You can achieve what you want with:
var myIdPostFix = myId.split("--").pop();