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Using regex to validate given password has alphabetic, numeric, and special characters not

I am trying to use regex in JavaScript to verify if a given password has alphabetic, numeric, and a special character. However, everything I have tried doesn't work
I have tried using
/^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!#"#%$&])[A-Za-z0-9!#"#%$&]{8,30}$/gm
and creating separate regex variables for alphabetic, numeric, and special characters:
let alpha = /^[A-Za-z]+$/i
let numer = /^[0-9]+$/i
let special = /^[!##$%^&*(),.?;":{}|<>']/i
Picture of My Code
When I log the password.match(regex) to the console I always see null

Try this:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[^\da-zA-Z]).{8,15}$
This is from a post a long time ago. There are probably others as well. A simple search of SO will find more examples.1
Here is a visualisation of the regular expression:
Debuggex Demo

I would personally go for separate regexes, the issue with your current regexes is that you have ^ at the start and $ at the end. Which means that the password must only contain [A-Za-z] from start to finish. Then you check if the password only contains [0-9] from start to finish.
const regexes = {
alpha: /[A-Za-z]/,
number: /[0-9]/,
special: /[!##$%^&*(),.?;":{}|<>']/,
length: /^.{8,30}$/
};
["dgXHUYuDdp", "zMv4qQfZj3", "4JXyrsq!J0", "a5Z!"].forEach(password => {
let valid = Object.values(regexes).every(regex => password.match(regex));
console.log(
"password: " + JSON.stringify(password) + "\n" +
"valid: " + valid
);
});

While I suspect it may be possible to write a regex which will do position/seuqnce independent matching, my head hurts just thinking about. So even if I could work out a way of doing it, I would not implement it - code needs to be readable and parseable by human beings. Looking at what you have presented here, I think I'm a lot more familiar with regexes than you are - so even more reason not to do this.
Your alpha / numer / special regexes will only match a string containing letters, numbers or special characters, not a mixture. If you change them thus, then you can check for a match of all three (and escape the meta characters in the special regex):
let alpha = /[A-Za-z]/i;
let numer = /[0-9]/;
let special = /[!##$%\^&*(),\.?;":{}|<>\']/;
if (password.match(alpha) && password.match(numer) && password.match(speicial)) {

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.

How to split a string by a character not directly preceded by a character of the same type?

Let's say I have a string: "We.need..to...split.asap". What I would like to do is to split the string by the delimiter ., but I only wish to split by the first . and include any recurring .s in the succeeding token.
Expected output:
["We", "need", ".to", "..split", "asap"]
In other languages, I know that this is possible with a look-behind /(?<!\.)\./ but Javascript unfortunately does not support such a feature.
I am curious to see your answers to this question. Perhaps there is a clever use of look-aheads that presently evades me?
I was considering reversing the string, then re-reversing the tokens, but that seems like too much work for what I am after... plus controversy: How do you reverse a string in place in JavaScript?
Thanks for the help!

Here's a variation of the answer by guest271314 that handles more than two consecutive delimiters:
var text = "We.need.to...split.asap";
var re = /(\.*[^.]+)\./;
var items = text.split(re).filter(function(val) { return val.length > 0; });
It uses the detail that if the split expression includes a capture group, the captured items are included in the returned array. These capture groups are actually the only thing we are interested in; the tokens are all empty strings, which we filter out.
EDIT: Unfortunately there's perhaps one slight bug with this. If the text to be split starts with a delimiter, that will be included in the first token. If that's an issue, it can be remedied with:
var re = /(?:^|(\.*[^.]+))\./;
var items = text.split(re).filter(function(val) { return !!val; });
(I think this regex is ugly and would welcome an improvement.)

You can do this without any lookaheads:
var subject = "We.need.to....split.asap";
var regex = /\.?(\.*[^.]+)/g;
var matches, output = [];
while(matches = regex.exec(subject)) {
output.push(matches[1]);
}
document.write(JSON.stringify(output));
It seemed like it'd work in one line, as it did on https://regex101.com/r/cO1dP3/1, but had to be expanded in the code above because the /g option by default prevents capturing groups from returning with .match (i.e. the correct data was in the capturing groups, but we couldn't immediately access them without doing the above).
See: JavaScript Regex Global Match Groups
An alternative solution with the original one liner (plus one line) is:
document.write(JSON.stringify(
"We.need.to....split.asap".match(/\.?(\.*[^.]+)/g)
.map(function(s) { return s.replace(/^\./, ''); })
));
Take your pick!

Note: This answer can't handle more than 2 consecutive delimiters, since it was written according to the example in the revision 1 of the question, which was not very clear about such cases.
var text = "We.need.to..split.asap";
// split "." if followed by "."
var res = text.split(/\.(?=\.)/).map(function(val, key) {
// if `val[0]` does not begin with "." split "."
// else split "." if not followed by "."
return val[0] !== "." ? val.split(/\./) : val.split(/\.(?!.*\.)/)
});
// concat arrays `res[0]` , `res[1]`
res = res[0].concat(res[1]);
document.write(JSON.stringify(res));

How to compare two Strings and get Different part

now I have two strings,
var str1 = "A10B1C101D11";
var str2 = "A1B22C101D110E1";
What I intend to do is to tell the difference between them, the result will look like
A10B1C101D11
A10 B22 C101 D110E1
It follows the same pattern, one character and a number. And if the character doesn't exist or the number is different between them, I will say they are different, and highlight the different part. Can regular expression do it or any other good solution? thanks in advance!

Let me start by stating that regexp might not be the best tool for this. As the strings have a simple format that you are aware of it will be faster and safer to parse the strings into tokens and then compare the tokens.
However you can do this with Regexp, although in javascript you are hampered by the lack of lookbehind.
The way to do this is to use negative lookahead to prevent matches that are included in the other string. However since javascript does not support lookbehind you might need to go search from both directions.
We do this by concatenating the strings, with a delimiter that we can test for.
If using '|' as a delimiter the regexp becomes;
/(\D\d*)(?=(?:\||\D.*\|))(?!.*\|(.*\d)?\1(\D|$))/g
To find the tokens in the second string that are not present in the first you do;
var bothstring=str2.concat("|",str1);
var re=/(\D\d*)(?=(?:\||\D.*\|))(?!.*\|(.*\d)?\1(\D|$))/g;
var match=re.exec(bothstring);
Subsequent calls to re.exec will return later matches. So you can iterate over them as in the following example;
while (match!=null){
alert("\""+match+"\" At position "+match.index);
match=re.exec(t);
}
As stated this gives tokens in str2 that are different in str1. To get the tokens in str1 that are different use the same code but change the order of str1 and str2 when you concatenate the strings.
The above code might not be safe if dealing with potentially dirty input. In particular it might misbehave if feed a string like "A100|A100", the first A100 will not be considered as having a missing object because the regexp is not aware that the source is supposed to be two different strings. If this is a potential issue then search for occurences of the delimiting character.

You call break the string into an array
var aStr1 = str1.split('');
var aStr2 = str2.split('');
Then check which one has more characters, and save the smaller number
var totalCharacters;
if(aStr1.length > aStr2.length) {
totalCharacters = aStr2.length
} else {
totalCharacters = aStr1.length
}
And loop comparing both
var diff = [];
for(var i = 0; i<totalCharacters; i++) {
if(aStr1[i] != aStr2[i]) {
diff.push(aStr1[i]); // or something else
}
}
At the very end you can concat those last characters from the bigger String (since they obviously are different from the other one).
Does it helps you?

Javascript string validation using the regex object

I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}

By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.

If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.