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Regex for any character except quote after comma

I want to match every word separated by comma, but it must not include a quote like ' or ".
I was using this regex:
^[a-zA-Z0-9][\!\[\#\\\:\;a-zA-Z0-9`_\s,]+[a-zA-Z0-9]$
However, it only matches a character and number and not a symbol.
The output should be:
example,example //true
exaplle,examp#3 //true, with symbol or number
example, //false, because there is no word after comma
,example //false, because there is no word before comma
##example&$123,&example& //true, with all character and symbol except quote

You can match 1+ times what is present in the character class. Then repeat 1+ times in a non capturing group (?: what is present in the character class, preceded by a comma.
^[!\[#\\:;a-zA-Z0-9`_ &$#]+(?:,[!\[#\\:;a-zA-Z0-9`_ &$#]+)+$
Regex demo
Note that you don't have to escape \!, \#, \: and \; in the character class, and that \s might also possibly match a newline.

I'm assuming you want the whole string to match perfectly with your conditions and return true then and then only.
These are the conditions-
Each word should be separated by a comma, said comma should have 2 valid words on each side
Words can contain anything except the 2 kinds of quotes (' and ") and whitespace characters (spaces and newlines).
The regex you would use is this- ^(?:[^,'"\s]+,[^,'"\s]+)+$, with the global flag (g) on.
Check out the demo here
Edit: As per request of being able to match only a single word.
This is the regex you would use for that- ^(?:(?:[^,'"\s]+,[^,'"\s]+)+|[^,'"\s]+)$
This will match words separated by a , as well as match just a single word.
The conditions for what qualifies as a word remains the same as aforementioned.
Quick explanation:-
^[^,'"\s]+,[^,'"\s]+$
This part matches 2 words separated by a comma, [^,'"\s]+ denotes a word
Wrapping that whole thing in ^(?:[^,'"\s]+,[^,'"\s]+)+$ simply makes it repeat, so it'll match N number of words separated by a comma, not just 2
Then adding another alternative using | and wrapping the whole thing in a group (non-capturing), we get ^(?:(?:[^,'"\s]+,[^,'"\s]+)+|[^,'"\s]+)$
This simply just adds the alternative [^,'"\s]+ - which matches a singular word.
Check out the updated demo here

Javascript: Regex to exclude whitespace and special characters

I need a regex to validate,
Should be of length 18
First 5 characters should be either (xyz34|xyz12)
Remaining 13 characters should be alphanumeric only letters and numbers, no whitespace or special characters is allowed.
I have a pattern like here, '/^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13}/g'
But this is allowing whitespace and special characters like ($,% and etc) which is violating the rule #3.
Any suggestion to exclude this whitespace and special characters and to strictly check that it must be letters and numbers?

You should not quantify lookarounds. They are non-consuming patterns, i.e. the consecutive positive lookaheads check the presence of their patterns but do not advance the regex index, they check the text at the same position. It makes no sense repeating them 13 times. ^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13} is equal to ^(xyz34|xyz12)(?=.*[a-zA-Z])(?=.*[0-9]), and means the string can start with xyz34 or xyz12 and then should have at least 1 letter and at least 1 digits.
You may consider fixing the issue by using a consuming pattern like this:
If you do not care if the last 13 chars contain only digits or only letters, use the patterns suggested by other users, like /^(?:xyz34|xyz12)[a-zA-Z\d]{13}$/ or /^xyz(?:34|12)[a-zA-Z0-9]{13}$/
If there must be at least 1 digit and at least 1 letter among those 13 alphanumeric chars, use /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/.
See the regex demo #1 and the regex demo #2.
NOTE: these are regex literals, do not use them inside single- or double quotes!
Details
^ - start of string
xyz - a common prefix
(?:34|12) - a non-capturing group matching 34 or 12
(?=[a-zA-Z]*\d) - there must be at least 1 digit after any 0+ letters to the right of the current location
(?=\d*[a-zA-Z]) - there must be at least 1 letter after any 0+ digtis to the right of the current location
[a-zA-Z\d]{13} - 13 letters or digits
$ - end of string.
JS demo:
var strs = ['xyz34abcdefghijkl1','xyz341bcdefghijklm','xyz34abcdefghijklm','xyz341234567890123','xyz14a234567890123'];
var rx = /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/;
for (var s of strs) {
console.log(s, "=>", rx.test(s));
}

.* will match any string, for your requirment you can use this:
/^xyz(34|12)[a-zA-Z0-9]{13}$/g
regex fiddle

/^(xyz34|xyz12)[a-zA-Z0-9]{13}$/
This should work,
^ asserts position at the start of a line
1st Capturing Group (xyz34|xyz12)
1st Alternative xyz34 matches the characters xyz34 literally (case sensitive)
2nd Alternative xyz12 matches the characters xyz12 literally (case sensitive)
Match a single character present in the list below [a-zA-Z0-9]{13}
{13} Quantifier — Matches exactly 13 times

Issue with javascript regex not matching less than 3 characters

I have the following javascript regex:
/^[^\s][a-z0-9 ]+[^\s]$/i
I need to allow any alphanumeric character as well as spaces inside the string but not at the beginning nor at the end.
Oddly enough, the above regex will not accept less than 3 characters, e.g. aa will not match but aaa will.
I am not sure why. Can anyone please help ?

You have: [^\s] (requires matching at least one non-whitespace character), [a-z0-9 ]+ (requires matching at least one alphanumeric or space character), and [^\s] again (requires matching at least one non-whitespace character). So, in total, you need at least 3 characters in the string.
Use word boundaries at the beginning and end instead:
/^\b[a-z0-9 ]+\b$/i
https://regex101.com/r/2GhH3N/1

Try the following regex:
^(?! )[a-z0-9 ]*[a-z0-9]$
Details:
^(?! ) - Start of the string and no space after it (so here we exclude the
initial space).
[a-z0-9 ]* - A sequence of letters, digits and spaces, possibly empty
(the content before the last letter(see below).
[a-z0-9]$ - The last letter and the end of string (so here we exclude the
terminal space).

You should re-write the expression as
/^[a-z0-9]+(?:\s+[a-z0-9]+)*$/i
See the regex demo.
NOTE: If only one whitespace is allowed between the alphanumeric chars use
/^[a-z0-9]+(?:\s[a-z0-9]+)*$/i
^^
Details
^ - start of string
[a-z0-9]+ - 1+ letters/digits
(?:\s+[a-z0-9]+)* - 0 or more repetitions of 1+ whitespaces (\s+) and 1+ digit/letters
$ - end of string.
See the regex graph:

JavaScript RegEx for matching a string having length between 1 to 5 characters and should not start and end with space

Need help in creating a JavaScript RegEx for matching a string having length between 1 to 5 characters and should not start and end with space. But it can have spaces in between.
I can come up with following
/(^\S\w*){1,5}/
But it is allowing string of length greater than 5 characters and it is also allowing string with trailing space

Assuming m m (three spaces between the m’s) is valid and you you want to allow any non-whitespace characters (\S), not just word characters (\w), you can restate your requirements as:
One non-whitespace character, optionally followed by:
Zero to three characters of any kind
One non-whitespace character
(The key is to group those last two.)
That gives us:
^\S(?:.{0,3}\S)?$
Try it out. (Thanks to RobC for the original examples, which I expanded a bit.)
If you want to match only word characters, you can instead say:
^\w(?:[\s\w]{0,3}\w)?$
Try it out.

JavaScript regular expressions to match no digits, whitespace and selected symbols

Thanks for taking a look.
My goal is to come up with a regexp that will match input that contains no digits, whitespace or the symbols !#£$%^&*()+= or any other symbol I may choose.
I am however struggling to grasp precisely how regular expressions work.
I started out with the simple pattern /\D/, which from my understanding will match the first non-digit character it can find. This would match the string 'James' which is correct but also 'James1' which I don't want.
So, my understanding is that if I want to ensure that a pattern is not found anywhere in a given string, I use the ^ and $ characters, as in /^\D$/. Now because this will only match a single character that is not a digit, I needed to use + to specify that 1 or more digits should not be founds in the entire string, giving me the expression /^\D+$/. Brilliant, it no longer matches 'James1'.
Question 1
Is my reasoning up to this point correct?
The next requirement was to ensure no whitespace is in the given string. \s will match a single whitespace and [^\s] will match the first non-whitespace character. So, from my understanding I just had to add this to what I have already to match strings that contain no digits and no whitespace. Again, because [^\s] will only match a single non-white space character, I used + to match one or more whitespace characters, giving the new regexp of /^\D+[^\s]+$/.
This is where I got lost, as the expression now matches 'James1' or even 'James Smith25'. What? Massively confused at this point.
Question 2
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Question 3
How would I go about writing the regular expression I'm trying to solve?
While I am keen to solve the problem I am more interested in figuring where my understanding of regular expressions is lacking, so any explanations would be helpful.

Not quite; ^ and $ are actually "anchors" - they mean "start" and "end", it's actually a little more complicated, but you can consider them to mean the start and end of a line for now - look up the various modifiers on regular expressions if you're interested in learning more about this. Unfortunately ^ has an overloaded meaning; if used inside square brackets it means "not", which is the meaning you are already acquainted with. It's very important that you understand the difference between these two meanings and that the definition in your head actually applies only to character range matching!
Contributing further to your confusion is that \d means "a numerical digit" and \D means "not a numerical digit". Similarly \s means "a whitespace (space/tab/newline/etc.) character" and \S means "not a whitespace character."
It's worth noting that \d is effectively a shortcut for [0-9] (note that - has a special meaning inside square brackets), and \D is a shortcut for [^0-9].
The reason it's matching strings that contain spaces is that you've asked for "1+ non-numerical digits followed by 1+ non-space characters" - so it'll match lots of strings! I think that perhaps you don't understand that regular expressions match bits of strings, you're not adding constraints as you go, but rather building up bots of matchers that will match bits of corresponding strings.
/^[^\d\s!#£$%^&*()+=]+$/ is the answer you're looking for - I'd look at it like this:
i. [] - match a range of characters
ii. []+ - match one or more of that range of characters
iii. [^\d\s]+ - match one or more characters that do not match \d (numerical digit) or \s (whitespace)
iv. [^\d\s!#£$%^&*()+=]+ - here's a bunch of other characters I don't want you to match
v. ^[^\d\s!#£$%^&*()+=]+$ - now there are anchors applied, so this matcher has to apply to the whole line otherwise it fails to match
A useful website to explore regexs is http://regexr.com/3b9h7 - which I supply with my suggested solution as an example. Edit: Pruthvi Raj's link to debuggerx is awesome!

Is my reasoning up to this point correct?
Almost. /\D/ matches any character other than a digit, but not just the first one (if you use g option).
and [^\s] will match the first non-whitespace character
Almost, [^\s] will match any non-whitespace character, not just the first one (if you use g option).
/^\D+[^\s]+$/ matching strings that contain spaces?
Yes, it does, because \D matches a space (space is not a digit).
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Because \D+ in /^\D+[^\s]+$/can match spaces.
Conclusion:
Use
^[^\d\s!#£$%^&*()+=]+$
It will match strings that have no digits and spaces, and the symbols you do not allow.
Mind that to match a literal -, ] or [ with a character class, you either need to escape them, or use at the start or end of the expression. To play it safe, escape them.

Just insert every character you don't want to include in a negated character class as follows:
^[^\s\d!#£$%^&*()+=]*$
DEMO
Debuggex Demo
^ - start of the string
[^...] - matches one character that is not in `...`
\s - matches a whitespace (space, newline,tab)
\d - matches a digit from 0 to 9
* - a quantifier that repeats immediately preceeding element by 0 or more times
so the regex matches any string that has
1. string that has a beginning
2. containing 0 or more number of characters that is not whitesapce, digit, and all the symbols included in the character class ( In this example !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class