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Regular expression for an alphanumeric word start with alphabet

I have a requirement to find and return the first occurrence of the pattern from a string.
Example: Please find my model number RT21M6211SR/SS and save it
Expected output: RT21M6211SR/SS
Condition for the pattern to match
Combination of digits and alphabets
Character length between 6 to 14
May or may not contain special characters like '-' or '/'
Starts with always alphabet
What I tried, but it didn't work for 4th condition
var str = 'Please find my model number RT21M6211SR/SS and save it';
var reg = /\b(\w|\d)[\d|\w-\/]{6,14}\b/;
var extractedMNO = '';
var mg = str.match(reg) || [""];
console.log('regular match mno', mg[0]);

\w matches word characters, which includes _ and digits as well. If you only want to match alphabetical characters, use [a-z] to match the first character.
Also, because you want to match lengths of 6-14, after matching the first character, you should repeat the character set with {5,13}, so that the repeated characters plus the first character comes out to a length of 6-14 characters.
var str = 'Please find my model number RT21M6211SR/SS and save it';
console.log(str.match(/\b[a-z][a-z0-9\/-]{5,13}/gi)[2]);
But since the matched string must contain digits (and doesn't just permit digits), then you need to make sure a digit exists in the matched substring as well, which you can accomplish by using lookahead for a digit right after matching the alphabetical at the start:
var str = 'Please find my model number RT21M6211SR/SS and save it';
console.log(str.match(/\b[a-z](?=[a-z\/-]{0,12}[0-9])[a-z0-9\/-]{5,13}/gi));
// ^^^^^^^^^^^^^^^^^^^^^^^
If you want to permit other special characters, just add them to the character set(s).

Regex: match last and second-last character

I need to wrap the two last characters in a string in a separate <span>:
-1:23 // This is what I have
-1:<span>2</span><span>3</span> // This is what I want
The following matches the last character – but how can I make it match the second last as well?
str.replace(/(.$)/, "<span>$1</span>");
Thanks :)

You may use
.replace(/.(?=.?$)/g, "<span>$&</span>")
See the regex demo
If these must be digits, replace . with \d:
.replace(/\d(?=\d?$)/g, "<span>$&</span>")
The pattern matches
\d - a digit
(?=\d?$) - that is followed with an end of string or a digit and end of string.
The $& is a replacement backreference that references the whole match value from the string replacement pattern.
JS demo:
console.log("-1:23".replace(/.(?=.?$)/g, "<span>$&</span>"));
console.log("-1:23".replace(/\d(?=\d?$)/g, "<span>$&</span>"));
Now, to make it more dynamic, you may use a limiting (range/interval) quantifier:
function wrap_chars(text, num_chars) {
var reg = new RegExp(".(?=.{0," + (num_chars-1) + "}$)", "g");
return text.replace(reg, "<span>$&</span>");
}
console.log(wrap_chars("-1:23", 1)); // wrap one char at the end with span
console.log(wrap_chars("-1:23", 2)); // wrap two chars at the end with span

You can add another group before the last one, which also matches a single character ((.)), then wrap each of them using references ($1 and $2):
var str = '-1:23'.replace(/(.)(.)$/, '<span>$1</span><span>$2</span>')
console.log(str);

Javascript - Regex - how to filter characters that are not part of regex

I want to accept words and some special characters, so if my regex
does not fully match, let's say I display an error,
var re = /^[[:alnum:]\-_.&\s]+$/;
var string = 'this contains invalid chars like ##';
var valid = string.test(re);
but now I want to "filter" a phrase removing all characters not matching the regex ?
usualy one use replace, but how to list all characters not matching the regex ?
var validString = string.filter(re); // something similar to this
how do I do this ?
regards
Wiktor Stribiżew solution works fine :
regex=/[^a-zA-Z\-_.&\s]+/g;
let s='some bloody-test #rfdsfds';
s = s.replace(/[^\w\s.&-]+/g, '');
console.log(s);
Rajesh solution :
regex=/^[a-zA-Z\-_.&\s]+$/;
let s='some -test #rfdsfds';
s=s.split(' ').filter(x=> regex.test(x));
console.log(s);

JS regex engine does not support POSIX character classes like [:alnum:]. You may use [A-Za-z0-9] instead, but only to match ASCII letters and digits.
Your current regex matches the whole string that contains allowed chars, and it cannot be used to return the chars that are not matched with [^a-zA-Z0-9_.&\s-].
You may remove the unwanted chars with
var s = 'this contains invalid chars like ##';
var res = s.replace(/[^\w\s.&-]+/g, '');
var notallowedchars = s.match(/[^\w\s.&-]+/g);
console.log(res);
console.log(notallowedchars);
The /[^\w\s.&-]+/g pattern matches multiple occurrences (due to /g) of any one or more (due to +) chars other than word chars (digits, letters, _, matched with \w), whitespace (\s), ., & and -.

To match all characters that is not alphanumeric, or one of -_.& move ^ inside group []
var str = 'asd.=!_#$%^&*()564';
console.log(
str.match(/[^a-z0-9\-_.&\s]/gi),
str.replace(/[^a-z0-9\-_.&\s]/gi, '')
);

Javascript regex ending word not found

In this regex , I need to match an ending word that starts with '(', ')' or ','.
Regex :
/[\(\),].*$/
For example, given the text (aaa,bbb)ccc I need to obtain )ccc. Still, it returns the entire text. What's wrong with this regex?

You can use:
'(aaa,bbb)ccc'.match(/[(),][^(),]+$/)
//=> [")ccc"]
[^(),]+ is negation pattern that matches any character but any listed in [^(),].
Problem with [(),].*$ is that it matches very first ( in input and matches till end.

You can also consider using capturing group while consuming all the characters up to the first (, ) or ,:
.*([(),].*)$
.* will consume as many characters as it can, then any character in this character class [(),], and then the rest of the characters up to the end.
The )ccc value will be stored in Group 1:
var re = /.*([(),].*)$/;
var str = '(aaa,bbb)ccc';
if ((m = re.exec(str)) !== null) {
document.getElementById("res").innerHTML = m[1];
}
<div id="res"/>

Try this regex:
/[(,)][^(,)]*$/

regex string replace

I am trying to do a basic string replace using a regex expression, but the answers I have found do not seem to help - they are directly answering each persons unique requirement with little or no explanation.
I am using str = str.replace(/[^a-z0-9+]/g, ''); at the moment. But what I would like to do is allow all alphanumeric characters (a-z and 0-9) and also the '-' character.
Could you please answer this and explain how you concatenate expressions.

This should work :
str = str.replace(/[^a-z0-9-]/g, '');
Everything between the indicates what your are looking for
/ is here to delimit your pattern so you have one to start and one to end
[] indicates the pattern your are looking for on one specific character
^ indicates that you want every character NOT corresponding to what follows
a-z matches any character between 'a' and 'z' included
0-9 matches any digit between '0' and '9' included (meaning any digit)
- the '-' character
g at the end is a special parameter saying that you do not want you regex to stop on the first character matching your pattern but to continue on the whole string
Then your expression is delimited by / before and after.
So here you say "every character not being a letter, a digit or a '-' will be removed from the string".

Just change + to -:
str = str.replace(/[^a-z0-9-]/g, "");
You can read it as:
[^ ]: match NOT from the set
[^a-z0-9-]: match if not a-z, 0-9 or -
/ /g: do global match
More information:
https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions

Your character class (the part in the square brackets) is saying that you want to match anything except 0-9 and a-z and +. You aren't explicit about how many a-z or 0-9 you want to match, but I assume the + means you want to replace strings of at least one alphanumeric character. It should read instead:
str = str.replace(/[^-a-z0-9]+/g, "");
Also, if you need to match upper-case letters along with lower case, you should use:
str = str.replace(/[^-a-zA-Z0-9]+/g, "");

str = str.replace(/\W/g, "");
This will be a shorter form

We can use /[a-zA-Z]/g to select small letter and caps letter sting in the word or sentence and replace.
var str = 'MM-DD-yyyy'
var modifiedStr = str.replace(/[a-zA-Z]/g, '_')
console.log(modifiedStr)