Develop Reference

Why would the replace with regex not work even though the regex does?

There may be a very simple answer to this, probably because of my familiarity (or possibly lack thereof) of the replace method and how it works with regex.
Let's say I have the following string: abcdefHellowxyz
I just want to strip the first six characters and the last four, to return Hello, using regex... Yes, I know there may be other ways, but I'm trying to explore the boundaries of what these methods are capable of doing...
Anyway, I've tinkered on http://regex101.com and got the following Regex worked out:
/^(.{6}).+(.{4})$/
Which seems to pass the string well and shows that abcdef is captured as group 1, and wxyz captured as group 2. But when I try to run the following:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
to replace those captured groups with "" I receive an empty string as my final output... Am I doing something wrong with this syntax? And if so, how does one correct it, keeping my original stance on wanting to use Regex in this manner...
Thanks so much everyone in advance...

The code below works well as you wish
"abcdefHellowxyz".replace(/^.{6}(.+).{4}$/,"$1")
I think that only use ()to capture the text you want, and in the second parameter of replace(), you can use $1 $2 ... to represent the group1 group2.
Also you can pass a function to the second parameter of replace,and transform the captured text to whatever you want in this function.
For more detail, as #Akxe recommend , you can find document on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace.

You are replacing any substring that matches /^(.{6}).+(.{4})$/, with this line of code:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
The regex matches the whole string "abcdefHellowxyz"; thus, the whole string is replaced. Instead, if you are strictly stripping by the lengths of the extraneous substrings, you could simply use substring or substr.
Edit
The answer you're probably looking for is capturing the middle token, instead of the outer ones:
var str = "abcdefHellowxyz";
var matches = str.match(/^.{6}(.+).{4}$/);
str = matches[1]; // index 0 is entire match
console.log(str);

Match a word unless it is preceded by an equals sign?

I have the following string
class=use><em>use</em>
that when searched using us I want to transform into
class=use><em><b>us</b>e</em>
I've tried looking at relating answers but I can't quite get it working the way I want it to. I'm especially interested in this answer's callback approach.
Help appreciated

This is a good exercise for writing regular expressions, and here's a possible solution.
"useclass=use><em>use</em>".replace(/([^=]|^)(us)/g, "$1<b>$2</b>");
// returns "<b>us</b>eclass=use><em><b>us</b>e</em>"
([^=]|^) ensures that the prefix of any matched us is either not an equal sign, or it's the start of the string.
As #jamiec pointed out in the comments, if you are using this to parse/modify HTML, just stop right now. It's mathematically impossible to parse a CFG with a regular grammar (even with enhanced JS regexps you will have a bad time trying to achieve that.)

If you can make any assumptions about the structure of your document, you may be better off using an approach that operates on DOM elements directly rather than parsing the whole document with a regex.
Parsing HTML with a regex has certain problems that can be painful to deal with.
var element = document.querySelector('em');
element.innerHTML = element.innerHTML.replace('us', '<b>us</b>');
<div class=use><em>use</em>
</div>

I would first look for any character other than the equals sign [^=] and separate it by parentheses so that I can use it again in my replacement. Then another set of parentheses around the two characters us ought to do it:
var re = /([^=]|^)(us)/
That will give you two capture groups to work with (inside the parentheses), which you can represent with $1 and $2 in your replacement string.
str.replace( /([^=|^])(us)/, '$1<b>$2</b>' );

Regex to find web addresses in short copy

Having a short copy I need to match all occurrences of links to websites. To keep things simple a need to find out addresses in this format:
www.aaaaaa.bbbbbb
http://aaaaaa.bbbb
https://aa.bbbb
but also I need to take care of longer www/http/https versions:
www.aaaaa.bbbb.ccc.ddd.eeee
etc. So basically number of subdomains is not known. Now I came up with this regex:
(www\.([a-zA-Z0-9-_]|\.(?!\s))+)[\s|,|$]|(http(s)?:\/\/(?!\.)([a-zA-Z0-9-_]|\.(?!\s))+)[\s|,|$]
If you test on:
this is some tex with www.somewIebsite.dfd.jhh.hjh inside of it or maybe http://www.ssss.com or maybe https://evenore.com hahaah blah
It works fine with exception of when address is at the very end. $ seems to work only when there is \n in the end and it fails for:
this is some tex with www.somewIebsite.dfd.jhh.hjh
I'm guessing fix is simple and I miss something obvious so how would I fix it? BTW I posted regex here if yu want to quickly play around https://regex101.com/r/eL1bI4/3

The problem is that you placed the end anchor $ inside the character group []
[\s|,|$]
It is then interpreted literally as a dollar sign, and not as the anchor (the pipe character | is also interpreted literally, it's not needed there). The solution is to move the $ anchor outside:
(?:[\s,]|$)
However, in this case it makes more sense to use a positive lookahead instead of the noncapturing group (you don't want trailing spaces, or commas):
(?=[\s,]|$)
In the result you will end up with the following regex pattern:
(www\.([a-zA-Z0-9-_]|\.(?!\s))+)(?=[\s,]|$)|(http(s)?:\/\/(?!\.)([a-zA-Z0-9-_]|\.(?!\s))+)(?=[\s,]|$)
See the working demo.
The updated version that handles trailing full stops:
(www\.([a-zA-Z0-9-_]|\.(?!\s|\.|$))+)(?=[\s,.]|$)|(http(s)?:\/\/(?!\.)([a-zA-Z0-9-_]|\.(?!\s|\.|$))+)(?=[\s,.]|$)
See the working demo.

Breaking a String into Chunks based on Pattern

I have one string, that looks like this:
a[abcdefghi,2,3,jklmnopqr]
The beginning "a" is fixed and non-changing, however the content within the brackets is and can follow a pattern. It will always be an alphabetical string, possibly followed by numbers separate by commas or more strings and/or numbers.
I'd like to be able to break it into chunks of the string and any numbers that follow it until the "]" or another string is met.
Probably best explained through examples and expected ideal results:
a[abcdefghi] -> "abcdefghi"
a[abcdefghi,2] -> "abcdefghi,2"
a[abcdefghi,2,3,jklmnopqr] -> "abcdefghi,2,3" and "jklmnopqr"
a[abcdefghi,2,3,jklmnopqr,stuvwxyz] -> "abcdefghi,2,3" and "jklmnopqr" and "stuvwxyz"
a[abcdefghi,2,3,jklmnopqr,1,9,stuvwxyz] -> "abcdefghi,2,3" and "jklmnopqr,1,9" and "stuvwxyz"
a[abcdefghi,1,jklmnopqr,2,stuvwxyz,3,4] -> "abcdefghi,1" and "jklmnopqr,2" and "stuvwxyz,3,4"
Ideally a malformed string would be partially caught (but this is a nice extra):
a[2,3,jklmnopqr,1,9,stuvwxyz] -> "jklmnopqr,1,9" and "stuvwxyz"
I'm using Javascript and I realize a regex won't bring me all the way to the solution I'd like but it could be a big help. The alternative is to do a lot of manually string parsing which I can do but doesn't seem like the best answer.
Advice, tips appreciated.
UPDATE: Yes I did mean alphametcial (A-Za-z) instead of alphanumeric. Edited to reflect that. Thanks for letting me know.

You'd probably want to do this in 2 steps. First, match against:
a\[([^[\]]*)\]
and extract group 1. That'll be the stuff in the square brackets.
Next, repeatedly match against:
[a-z]+(,[0-9]+)*
That'll match things like "abcdefghi,2,3". After the first match you'll need to see if the next character is a comma and if so skip over it. (BTW: if you really meant alphanumeric rather than alphabetic like your examples, use [a-z0-9]*[a-z][a-z0-9]* instead of [a-z]+.)
Alternatively, split the string on commas and reassemble into your word with number groups.

Why wouldn't a regex bring you all the way to a solution?
The following regex works against the given data, but it makes a few assumptions (at least two alphas followed by comma separated single digits).
([a-z]{2,}(?:,\\d)*)
Example:
re = new RegExp('[a-z]{2,}(?:,\\d)*', 'g')
matches = re.exec("a[abcdefghi,2,3,jklmnopqr,1,9,stuvwxyz]")

Assuming you can easily break out the string between the brackets, something like this might be what you're after:
> re = new RegExp('[a-z]+(?:,\\d)*(?:,?)', 'gi')
> while (match = re.exec("abcdefghi,2,3,jklmnopqr,1,9,stuvwxyz")) { print(match[0]) }
abcdefghi,2,3,
jklmnopqr,1,9,
stuvwxyz
This has the advantage of working partially in your malformed case:
> while (match = re.exec("abcdefghi,2,3,jklmnopqr,1,9,stuvwxyz")) { print(match[0]) }
jklmnopqr,1,9,
stuvwxy
The first character class [a-z] can be modified if you meant for it to be truly alphanumeric.

How to search csv string and return a match by using a Javascript regex

I'm trying to extract the first user-right from semicolon separated string which matches a pattern.
Users rights are stored in format:
LAA;LA_1;LA_2;LE_3;
String is empty if user does not have any rights.
My best solution so far is to use the following regex in regex.replace statement:
.*?;(LA_[^;]*)?.*
(The question mark at the end of group is for the purpose of matching the whole line in case user has not the right and replace it with empty string to signal that she doesn't have it.)
However, it doesn't work correctly in case the searched right is in the first position:
LA_1;LA_2;LE_3;
It is easy to fix it by just adding a semicolon at the beginning of line before regex replace but my question is, why doesn't the following regex match it?
.*?(?:(?:^|;)(LA_[^;]*))?.*
I have tried numerous other regular expressions to find the solution but so far without success.

I am not sure I get your question right, but in regards to the regular expressions you are using, you are overcomplicating them for no clear reason (at least not to me). You might want something like:
function getFirstRight(rights) {
var m = rights.match(/(^|;)(LA_[^;]*)/)
return m ? m[2] : "";
}

You could just split the string first:
function getFirstRight(rights)
{
return rights.split(";",1)[0] || "";
}

To answer the specific question "why doesn't the following regex match it?", one problem is the mix of this at the beginning:
.*?
eventually followed by:
^|;
Which might be like saying, skip over any extra characters until you reach either the start or a semicolon. But you can't skip over anything and then later arrive at the start (unless it involves newlines in a multiline string).
Something like this works:
.*?(\bLA_[^;]).*
Meaning, skip over characters until a word boundary followed by "LA_".