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How to replace string between two string with the same length

I have an input string like this:
ABCDEFG[HIJKLMN]OPQRSTUVWXYZ
How can I replace each character in the string between the [] with an X (resulting in the same number of Xs as there were characters)?
For example, with the input above, I would like an output of:
ABCDEFG[XXXXXXX]OPQRSTUVWXYZ
I am using JavaScript's RegEx for this and would prefer if answers could be an implementation that does this using JavaScript's RegEx Replace function.
I am new to RegEx so please explain what you do and (if possible) link articles to where I can get further help.

Using replace() and passing the match to a function as parameter, and then Array(m.length).join("X") to generate the X's needed:
var str = "ABCDEFG[HIJKLMN]OPQRSTUVWXYZ"
str = str.replace(/\[[A-Z]*\]/g,(m)=>"["+Array(m.length-1).join("X")+"]")
console.log(str);
We could use also .* instead of [A-Z] in the regex to match any character.
About regular expressions there are thousands of resources, specifically in JavaScript, you could see Regular Expressions MDN but the best way to learn, in my opinion, is practicing, I find regex101 useful.

const str="ABCDEFG[HIJKLMN]OPQRSTUVWXYZ";
const run=str=>str.replace(/\[.*]/,(a,b,c)=>c=a.replace(/[^\[\]]/g,x=>x="X"));
console.log(run(str));
The first pattern /\[.*]/ is to select letters inside bracket [] and the second pattern /[^\[\]]/ is to replace the letters to "X"

We can observe that every individual letter you wish to match is followed by a series of zero or more non-'[' characters, until a ']' is found. This is quite simple to express in JavaScript-friendly regex:
/[A-Z](?=[^\[]*\])/g
regex101 example
(?= ) is a "positive lookahead assertion"; it peeks ahead of the current matching point, without consuming characters, to verify its contents are matched. In this case, "[^[]*]" matches exactly what I described above.
Now you can substitute each [A-Z] matched with a single 'X'.

You can use the following solution to replace a string between two square brackets:
const rxp = /\[.*?\]/g;
"ABCDEFG[HIJKLMN]OPQRSTUVWXYZ".replace(rxp, (x) => {
return x.replace(rxp, "X".repeat(x.length)-2);
});

Find number that follows certain string which includes both letters and punctuation

I am trying to find a way to extract the numbers that occur after abc/ immediately succeeding the / and before any further letters, numbers or punctuation.
E.g:
abc/134567/something should return 1234567
abc/1234567?foo=bar should still only return 1234567
blah/1234/abc/678 should only return 678 as I'm looking only for the number that succeeds abc/
I'm aware there are two options: regex or substring match.
In order to perform the substring match I need the index point but I'm dubious about merely doing an indexOf("abc/") as it only returns the index of the first letter - a - which could be present elsewhere in the string.
With regex I have struggled as I find that searching for a mixture of the letters and the slashes seems to cause it to return null.
So what's the best way?

You can use this regexpression :
var rgx = new RegExp("abc\/([0-9]+)","gi");
Then :
var m = rgx.exec("abc/1234567?foo=bar");
console.log(m[0]);
edited after comments

You could use a regular expression and seach for abc/ and following digits.
var array = ['abc/134567/something', 'abc/1234567?foo=bar', 'blah/1234/abc/678'];
console.log(array.map(s => s.match(/abc\/(\d+)/)[1]));

We accept string that has abc/, after it an integer number, that is taken as a matched group and either the end of string or some non-digit symbol after it.
abc\/(\d+)(?:$|\D)
test
You'll use in Javascript for matched group extraction:
var myRegexp = /abc\/(\d+)(?:$|\D)/g;
var match = myRegexp.exec(inputString);
var result=match[1]; // the number after abc/
In another regex engine than that of JavaScript, lookahead and lookbehind could be used. But in JS lookbehinds are forbidden. :-(. So we have to use this, a bit more complicated, way.

Are you after something like this:
^(.*\/)(\d+)(.*)
Where the second group will give you the digits after the slash.
Look at the regex here

JQuery match with RegEx not working

I have a filename that will be something along the lines of this:
Annual-GDS-Valuation-30th-Dec-2016-082564K.docx
It will contain 5 numbers followed by a single letter, but it may be in a different position in the file name. The leading zero may or may not be there, but it is not required.
This is the code I come up with after checking examples, however SelectedFileClientID is always null
var SelectedFileClientID = files.match(/^d{5}\[a-zA-Z]{1}$/);
I'm not sure what is it I am doing wrong.
Edit:
The 0 has nothing to do with the code I am trying to extract. It may or may not be there, and it could even be a completely different character, or more than one, but has nothing to do with it at all. The client has decided they want to put additional characters there.

There are at least 3 issues with your regex: 1) the pattern is enclosed with anchors, and thus requires a full string match, 2) the d matches a letter d, not a digit, you need \d to match a digit, 3) a \[ matches a literal [, so the character class is ruined.
Use
/\d{5}[a-zA-Z]/
Details:
\d{5} - 5 digits
[a-zA-Z] - an ASCII letter
JS demo:
var s = 'Annual-GDS-Valuation-30th-Dec-2016-082564K.docx';
var m = s.match(/\d{5}[a-zA-Z]/);
console.log(m[0]);

All right, there are a few things wrong...
var matches = files.match(/\-0?(\d{5}[a-zA-Z])\.[a-z]{3,}$/);
var SelectedFileClientID = matches ? matches[1] : '';
So:
First, I get the matches on your string -- .match()
Then, your file name will not start with the digits - so drop the ^
You had forgotten the backslash for digits: \d
Do not backslash your square bracket - it's here used as a regular expression token
no need for the {1} for your letters: the square bracket content is enough as it will match one, and only one letter.
Hope this helps!

Try this pattern , \d{5}[a-zA-Z]

Try - 0?\d{5}[azA-Z]
As you mentioned 0 may or may not be there. so 0? will take that into account.
Alternatively it can be done like this. which can match any random character.
(\w+|\W+|\d+)?\d{5}[azA-Z]

Regex - must contain number and must not contain special character

I want to check by regex if:
String contains number
String does not contain special characters (!<>?=+#{}_$%)
Now it looks like:
^[^!<>?=+#{}_$%]+$
How should I edit this regex to check if there is number anywhere in the string (it must contain it)?

you can add [0-9]+ or \d+ into your regex, like this:
^[^!<>?=+#{}_$%]*[0-9]+[^!<>?=+#{}_$%]*$
or
^[^!<>?=+#{}_$%]*\d+[^!<>?=+#{}_$%]*$
different between [0-9] and \d see here

Just look ahead for the digit:
var re = /^(?=.*\d)[^!<>?=+#{}_$%]+$/;
console.log(re.test('bob'));
console.log(re.test('bob1'));
console.log(re.test('bob#'))
The (?=.*\d) part is the lookahead for a single digit somewhere in the input.

You only needed to add the number check, is that right? You can do it like so:
/^(?=.*\d)[^!<>?=+#{}_$%]+$/
We do a lookahead (like peeking at the following characters without moving where we are in the string) to check to see if there is at least one number anywhere in the string. Then we do our normal check to see if none of the characters are those symbols, moving through the string as we go.
Just as a note: If you want to match newlines (a.k.a. line breaks), then you can change the dot . into [\W\w]. This matches any character whatsoever. You can do this in a number of ways, but they're all pretty much as clunky as each other, so it's up to you.

Javascript regex to check if a value begins with a number followed by anything else

How would i go about doing a regex to see if it begins with a number and any character can follow after. My current expression is
var validaddress = /^[0-9][A-Za-z0-9]+$/;
But this isn't the right way. Im new to this, help anyone?

If you need character(s) after the digit, try this:
var validaddress = /^[0-9].+$/;
If characters after the digit are optional, use this:
var validaddress = /^[0-9].*$/;

What you looking for is: var validaddress = /^\d.*$/;
\d - Matches any digit
.* - Matches any character except newline zero or more times.
Or replace .* with .+, if you are looking for at least 1 character.

Try /^[0-9]/ as the regular expression.
If it only needs to start with a number, I'd only check that...

when you say "any character follow" -- do you mean any alphanumeric character or just anything (i.e. including space, comma, slash etc)?
if it is the latter, how about this:
var validaddress = /^[0-9].+$/;

I suppose this should work
/^\d+.+$/

I would get rid of the $. Also, a '.' would suffice for "any character". This one works for me:
var validaddress = ^[0-9].+;

If you have a string of these values and you want to find each individually try this:
(^|(?<=\W))(\d\w*)
You can then do a loop through each match.
Regexr example

You could also use /^\d/ as the briefest approach.