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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

Regex: string up to 20char long, without specific characters

I am trying to make regexp for validating string not containing
^ ; , & . < > | and having 1-20 characters. Any other Unicode characters are valid (asian letters for example).
How to do it?

You can use the following:
^[^^;,&.<>|]{1,20}$
Explanation:
^ assert starting of the string
[^ start of negated character class ([^ ])
^;,&.<>| all the characters you dont want to match
] close the negates character class
{1,20} range of matches
$ assert ending of the string
It will match any character other than specified characters within range of 1-20.

Your regex \w[^;,&.<>|]{1,20} contains \w that might not match all Unicode letters (I guess your regex flavor does not match Unicode letters with \w). Anyway, the \w only matches 1 character in your pattern.
Also, you say you need to exclude ^ but it is missing in your pattern.
When you want to validate length, you also must use ^/$ anchors to mark the beginning and end of a string.
To create a pattern for some range that does not match specific characters, you need a negated character class with anchors around it, and the length is set with limiting quantifiers:
^[^^;,&.<>|]{1,20}$
Or (this version makes sure we only match at the beginning and end of the string, never a line):
\A[^^;,&.<>|]{1,20}\z
Note that inside a character class, almost all special characters do not require escaping (only some of them, none in your case). Even the ^ caret symbol.
See demo

Javascript regex - no white space at beginning + allow space in the middle

I would like to have a regex which matches the string with NO whitespace(s) at the beginning. But the string containing whitespace(s) in the middle CAN match. So far i have tried below
[^-\s][a-zA-Z0-9-_\\s]+$
Debuggex Demo
Above is not allowing whitespace(s) at the beginning, but also not allowing in the middle/end of the string. Please help me.

In your 2nd character class, \\s will match \ and s, and not \s. Thus it doesn't matches a whitespace. You should use just \s there. Also, move the hyphen towards the end, else it will create unintentional range in character class:
^[^-\s][a-zA-Z0-9_\s-]+$

If you plan to match a string of any length (even an empty string) that matches your pattern and does not start with a whitespace, use (?!\s) right after ^:
/^(?!\s)[a-zA-Z0-9_\s-]*$/
^^^^^^
Or, bearing in mind that [A-Za-z0-9_] in JS regex is equal to \w:
/^(?!\s)[\w\s-]*$/
The (?!\s) is a negative lookahead that matches a location in string that is not immediately followed with a whitespace (matched with the \s pattern).
If you want to add more "forbidden" chars at the string start (it looks like you also disallow -) keep using the [\s-] character class in the lookahead:
/^(?![\s-])[\w\s-]*$/
To match at least 1 character, replace * with +:
/^(?![\s-])[\w\s-]+$/
See the regex demo. JS demo:
console.log(/^(?![\s-])[\w\s-]+$/.test("abc def 123 ___ -loc- "));
console.log(/^(?![\s-])[\w\s-]+$/.test(" abc def 123 ___ -loc- "));

You need to use this regex:
^[^-\s][\w\s-]+$
Use start anchor ^
No need to double escape \s
Also important is to use hyphen as the first OR last character in the character class.
\w is same as [a-zA-Z0-9_]

use \S at the beginning
^\S+[a-zA-Z0-9-_\\s]+$

This RegEx will allow neither white-space at the beginning nor at the end of. Your string/word and allows all the special characters.
^[^\s].+[^\s]$
This Regex also works Fine
^[^\s]+(\s+[^\s]+)*$

try this should work
[a-zA-Z0-9_]+.*$

/^[^.\s]/
try this instead it will not allow a user to enter character at first place
^ matches position just before the first character of the string
. matches a single character. Does not matter what character it is, except newline
\s is space

If your field for user name only accept letters and middle of space but not for begining and end
User name: /^[^\s][a-zA-Z\s]+[^\s]$/
If your field for user ID only accept letters,numbers and underscore and no spaces allow
user ID: /^[\w]+$/
If your field for password only accept letters,number and special character no spaces allow
Password: /^[\w##&]+$/
Note: \w content a-zA-Z, number, underscore (_) if you add more character, add you special character after \w.
You can compare with user ID and password field in password field im only add some special character (##&).
India public thoko like 😁

I suggest below regex for this,
^[^\s].*[^\s]$
You can try regex in here

RegExp extract specific string followed by any number with leading / trailing whitespace

I want to extract a string from another using JavaScript / RegExp.
Here is what I got:
var string = "wp-button wp-image-45 wp-label";
string.match(/(?:(?:.*)?\s+)?(wp-image-([0-9]+))(:?\s(?:.*)?)?/);
// returnes: ["wp-button ", "wp-image-45", "45", undefined]
I just want to have "wp-image-45", so:
(Optional) any character
(Optional) followed by whitespace
(Required) followed by "wp-image-"
(Required) followed by any number
(Optional) followed by whitespacy
(Optional) followed by any character
What is missing here? Is it just some kind of bracketing or more?
I also tried
string.match(/(?:(?:.*)?\s+)?(?=(wp-image-([0-9]+)))(?=(:?\s(?:.*)?)?)/)
Edit: In the end I just want to have the number. But I'd also make this step in between.

Regexps are not required to start matching at the beginning of the string, so your attempts to match whitespace and any character aren't necessary. Also, "any character" includes whitespace (except newlines in certain modes).
This should be all you need:
string.match(/\bwp-image-(\d+)\b/)
This will capture, for example, "wp-image-123" into matching group 0, and "123" into matching group 1.
\b means "word boundary", which ensures that you won't match "abcwp-image-123def". A word boundary is defined as any place where a non-word character is followed by a word character, or vice versa. A word character is consists of a letter, a number or an underscore.
Also, I used \d instead of [0-9] simply out of convenience. They have slightly different meaning (\d also matches characters considered numbers in other languages), but that won't make a difference in your case.

If all of that surrounding stuff is optional and all you want is the number then there's no point to matching for any of that stuff except for that "wp-image-" prefix, just do:
var string = "wp-button wp-image-45 wp-label";
string.match(/wp-image-([0-9]+)/);

How to make the below regex to accept any special character

/^[^ ]([\w- \.\\\/&#]+)[^ ]$/,
I have the above regex. I want to make sure it accepts all special characters but i don't want to specify the entire special character listsuch as [\w- \.\\\/&#!##$&]. How can we make sure the above regex accepts all special characters

[^\w\s] matches any non-alphanumeric and non-whitespace character.
\S matches any non-whitespace character.
. matches any character except newlines.
[\S\s] matches any character in a JavaScript regex.

Since you've got \w and a space in there already, you must want all of the ASCII characters except control characters. That would be:
[ -~]
...or any character whose code point is in the range U+0020 (space) to U+007E (tilde). But it looks like you want to make sure the first and last characters are not whitespace. In fact, looking at your previous question, I'll assume you want only letters or digits in those positions. This would work:
/^[A-Za-z0-9][ -~]*[A-Za-z0-9]$/
...but that requires the string to be at least two characters long. To allow for a single-character string, change it to this:
/^[A-Za-z0-9](?:[ -~]*[A-Za-z0-9])?$/
In other words, if there's only one character, it must be a letter or digit. If there are two or more characters, the first and last must letters or digits, while the rest can be any printing character--i.e., a letter, a digit, a "special" (punctuation) character, or a space.
Note that this only matches ASCII characters, not accented Latin letters like  or ë, or symbols from other alphabets or writing systems.

. matches any character except for newline.