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Regex creation to allow, disallow few characters

I am new to regex, i have this use case:
Allow characters, numbers.
Zero or one question mark allowed. (? - valid, consecutive question marks are not allowed (??)).
test-valid
?test - valid
??test- invalid
?test?test - valid
???test-invalid
test??test -invalid
Exlcude $ sign.
[a-zA-Z0-9?] - seems this doesn't work
Thanks.

Try the following regular expression: ^(?!.*\?\?)[a-zA-Z0-9?]+$
first we're using Negetive lookahead - which allows us to exclude any character which is followed by double question marks (Negetive lookahaed does not consume characters)
Since question mark has special meaning in regular expressions (Quantifier — Matches between zero and one times), each question mark is escaped using backslash.
The plus sign at the end is a Quantifier — Matches between one and unlimited times, as many times as possible
You can test it here

Your description can be broken down into the regex:
^(?:\??[a-zA-Z0-9])+\??$
You say characters and your description shows letters and numbers only, but it's possible \w (word characters) may be used instead - this includes underscore
It's between ^ and $ meaning the whole field must match (no partial matches, although if you want those you can remove this. The + means there must be at least one match (so empty string won't match). The capturing group ((\??[a-zA-Z0-9])) says I must either see a question mark followed by letters or just letters repeating many times, and the final question mark allows the string to end with a single question mark.
You probably don't want capturing groups here, so we can start that with ?: to prevent capture leading to:
^(?:\??[a-zA-Z0-9])+\??$
Matches
test
?test
?test?test
test?
Doesn't match
??test
???test
test??test
test??
<empty string>
?

Regex to reject names ending in non-letters except dot

I have written a regex that returns true or false depending on whether the text provided is a valid first/last name:
let letters = `a-zA-Z`;
letters += `àáâäãåąčćęèéêëėįìíîïłńòóôöõøùúûüųūÿýżźñçčšž`;
letters += `ÀÁÂÄÃÅĄĆČĖĘÈÉÊËÌÍÎÏĮŁŃÒÓÔÖÕØÙÚÛÜŲŪŸÝŻŹÑßÇŒÆČŠŽ∂ð`;
const re = new RegExp(`^[${letters}][${letters} ,.'’-]+[${letters}.]$`, 'u')
return(name.match(re));
So far, I'm able to ensure it only validates names that actually start with a letter and do not contain numerals or any special characters other than dot, hyphen, or comma. However, it still rejects names like Jo and Xi. I understand it's due to the three separate $-blocks. But the blocks are there to ensure the name doesn't start with a non-letter or end in a non-letter other than dot. How should I modify my expression to accommodate this?
Also, is there any way to shorten this expression without compromising its range? I still need it to cover extended Latin characters.

If the minimum length of the word is 2 chars, you could use a negative lookahead ^(?!.*[ ,'’]$) to assert that the string does not end with the characters that you would not allow and leave out the last [${letters}.]
Regex demo
If the minimum length is 1, you could use another negative lookahead (?![ .,'’]) and add the dot as well so that a single dot is not allowed at the beginning and then use the single character class that contains all allowed characters.
^(?!.*[ ,'’]$)(?![ .,'’])[a-zA-ZàáâäãåąčćęèéêëėįìíîïłńòóôöõøùúûüųūÿýżźñçčšžÀÁÂÄÃÅĄĆČĖĘÈÉÊËÌÍÎÏĮŁŃÒÓÔÖÕØÙÚÛÜŲŪŸÝŻŹÑßÇŒÆČŠŽ∂ð ,.'’-]+$
Regex demo

Regex pattern for a String contains "at least one digit and allows only two special chars"

I have requirement like this for Regex pattern for a string:
1.String length >=8;
2.Contains at-least one digit.
3.Contains exactly two special characters in string.
4.remaining characters are alphabets.
I have tried like this:
"/^(?=.*[0-9]+)(?=.*[##$%]{2})[0-9##$%A-Za-z]{8,}$/g"
But in this examples i am getting:
1."Example1##" --true (passed my test)
2."Example2#" --false (passed my test)
3."Example3####$#" --true (failed my test)
==>In 3rd case it is accepting more than 2 special characters.
How to achieve my requirement, please help me solve this.

You can use
^(?=\D*\d)(?=(?:\w*[##$%]){2}(?!\w*[##$%]))[\d##$%A-Za-z]{8,}
https://regex101.com/r/9iyoIX/2
(?=\D*\d) - at least one digit
(?=(?:\w*[##$%]){2}(?!\w*[##$%])) - Repeat (optional word characters followed by a special character) 2 times, then negative lookahead for more special characters
(?=[\d##$%A-Za-z]{8,}) - String is at least 8 characters long, and only contains the desired types of characters
Note that \d is preferable to [0-9] - easier to read. Also, when you're looking for a certain number of characters of a particular type in a string, a negated character set has better performance than .*. For example, when looking for a digit, better to use
(?=\D*\d)
than
(?=.*\d)
which will be more efficient - the regex will fail quicker when there's no match.

Regex - how to ignore order of the matched groups? [duplicate]

This question already has answers here:
Password REGEX with min 6 chars, at least one letter and one number and may contain special characters
(10 answers)
Closed 2 years ago.
I'm trying to create a regex validation for a password which is meant to be:
6+ characters long
Has at least one a-z
Has at least one A-Z
Has at leat one 0-9
So, in other words, the match will have :
at least one a-z, A-Z, 0-9
at least 3 any other characters
I've came up with:
((.*){3,}[a-z]{1,}[A-Z]{1,}[0-9]{1,})
it seems pretty simple and logical to me, but 2 things go wrong:
quantifier {3,} for (.*) somehow doesn't work and destroys whole regex. At first I had {6,} at the end but then regex would affect the quantifiers in inner groups, so it will require [A-Z]{6,} instead of [A-Z]{1,}
when I remove {3,} the regex works, but will match only if the groups are in order - so that it will match aaBB11, but not BBaa11

This is a use case where I wouldn't use a single regular expression, but multiple simpler ones.
Still, to answer your question: If you only want to validate that the password matches those criteria, you could use lookaheads:
^(?=.{6})(?=.*?[a-z])(?=.*?[A-Z])(?=.*?[0-9])
You're basically looking for a position from which you look at
6 characters (and maybe more to follow, doesn't matter): (?=.{6})
maybe something, then a lowercase letter: (?=.*?[a-z])
maybe something, then an uppercase letter: (?=.*?[A-Z])
maybe something, then a digit: (?=.*?[0-9])
The order of appearance is arbitrary due to the maybe something parts.
(Note that I've interpreted 6 characters long as at least 6 characters long.)

I believe this is what you want:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])[!-~]{6,}$
If we follow your spec to the letter, your validation password looks like this:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).{6,}$
However, we need to improve on this, because apart from the number, lower-case and upper-case letter, are you really willing to accept any character? For instance, can the user use a character in the Thai language? A space character? A tab? Didn't think so. :)
If you want to allow all the printable ASCII characters apart from space, instead of a dot, we can use this character range: [!-~]
How does it work?
The ^ anchor makes sure we start the match at the start of the string
The (?=.*[a-z]) lookahead ensures we have a lower-case character
The (?=.*[A-Z]) lookahead ensures we have an upper-case character
The (?=.*[0-9]) lookahead ensures we a digit
The (?=.*[a-z]) lookahead ensures we have a lower-case character
The [!-~]{6,} matches six or more ASCII printable ASCII characters that are not space.
The $ ensures we have reached the end of the string (otherwise, the password could contain more characters that are not allowed).

you could use this pattern ^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).{6,}

Regex which accepts alphanumerics only, except for one hyphen in the middle

I am trying to construct a regular expression which accepts alphanumerics only ([a-zA-Z0-9]), except for a single hyphen (-) in the middle of the string, with a minimum of 9 characters and a maximum of 20 characters.
I have verified the following expression, which accepts a hyphen in the middle.
/^[a-zA-Z0-9]+\-?[a-zA-Z0-9]+$/
How can I set the minimum 9 and maximum 20 characters for the above regex? I have already used quantifiers + and ? in the above expression.
How would I apply {9,20} to the above expression? Are there any other suggestions for the expression?

/^[a-zA-Z0-9]+\-?[a-zA-Z0-9]+$/
can be simplified to
/^[a-z0-9]+(?:-[a-z0-9]+)?$/i
since if there is no dash then you don't need to look for more letters after it, and you can use the i flag to match case-insensitively and avoid having to reiterate both lower-case and upper-case letters.
Then split your problem into two cases:
9-20 alpha numerics
10-21 characters, all of which are alpha numerics except one dash
You can check the second using a positive lookahead like
/^(?=.{10,21}$)/i
to check the number of characters without consuming them.
Combining these together gives you
/^(?:[a-z0-9]{9,20}|(?=.{10,21}$)[a-z0-9]+-[a-z0-9]+)$/i

You can do this provided you don't want - to be present exactly in middle
/^(?=[^-]+-?[^-]+$)[a-zA-Z\d-]{9,20}$/
[^-] matches any character that is not -