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openapi - regex for not allowing whitespace or hyphen [duplicate]

I tried this but it doesn't work :
[^\s-]
Any Ideas?

[^\s-]
should work and so will
[^-\s]
[] : The char class
^ : Inside the char class ^ is the
negator when it appears in the beginning.
\s : short for a white space
- : a literal hyphen. A hyphen is a
meta char inside a char class but not
when it appears in the beginning or
at the end.

It can be done much easier:
\S which equals [^ \t\r\n\v\f]

Which programming language are you using? May be you just need to escape the backslash like "[^\\s-]"

In Java:
String regex = "[^-\\s]";
System.out.println("-".matches(regex)); // prints "false"
System.out.println(" ".matches(regex)); // prints "false"
System.out.println("+".matches(regex)); // prints "true"
The regex [^-\s] works as expected. [^\s-] also works.
See also
Regular expressions and escaping special characters
regular-expressions.info/Character class
Metacharacters Inside Character Classes
The hyphen can be included right after the opening bracket, or right before the closing bracket, or right after the negating caret.

Note that regex is not one standard, and each language implements its own based on what the library designers felt like. Take for instance the regex standard used by bash, documented here: https://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap09.html#tag_09_03_05.
If you are having problems with regular expressions not working, it might be good to simplify it, for instance using "[^ -]" if this covers all forms of whitespace in your case.

Try [^- ], \s will match 5 other characters beside the space (like tab, newline, formfeed, carriage return).

Why does this pattern not match? ([\\\\A\\\\W]its[\\\\W\\\\z])

I'm trying to do a replace with this pattern, so I need to match this:
String pattern = "[\\\\A\\\\W]its[\\\\W\\\\z]";
The way I'm interpreting my pattern is: either a beginning of the string OR a non word character like a space or comma, then an "its", then a non word character OR the end of the string.
Why doesn't it match on this "its" inside this string?
its about time
The idea of what this is supposed to do it's supposed to detect incorrectly written words like "its" and fix them to "it's".
Also why do I need so many escape characters in order for the pattern to be accepted by the vm at all?

\\A and \\z are boundary matches. They cannot go inside character classes. If you use them properly, i.e. with two slashes instead of four, regex pattern compiler would throw an exception, because \A or \z cannot go inside [] blocks.
Use straight | syntax with non-capturing groups instead:
String pattern = "(?:\\A|\\W)its(?:\\W|\\z)";
Demo.

How to create regular expression to match the end of word in Java?

Need regex for matching { character and every line that ends with this character. For example: (isMoving){
Tried:
^.?\{$
but it works only for single {

.? means zero or one character. You want to match zero or more. Also the curly brace is a special regex character that is used for quantifying the occurrences of a certain pattern, and so it should be escaped:
^.*\{$

The Java string representation of the regex you want is ".*\\{" for use like
if (lineText.matches(".*\\{")) { ... }
A good place to experiment with such things is RegexPlanet's Java page

REGEX: how to make this statement become 2 match in REGEX

I have statement: <%=anything%><%=anything%>
and a regular expression: <%=\\s*(\\S+)\\s*%>.
The regex matches the stament as 1 match instead of 2 matches.
Can someone fix my regex?
Btw I use Java for my application

You are currently matching it all into one match, because regex usually is greedy, thus taking everything it can match into the match - so =anything%><%=anything is all matched by \S+. You could use the lazy modifier for the \S, so it matches as small as it has to, like so: <%=\\s*(\\S+?)\\s*%>. But there is an even better way to work with - as you don't want to match the closing >, just include it into a negative character class: <%=\\s*([^\\s>]+)\\s*%>
Here is a demo of it: https://regex101.com/r/bA4qY9/1
Note that you might have to double the backslashes again after testing in regex101
If you want to read further into it, have a look at http://www.regular-expressions.info/repeat.html

Java String validation only one alphanumeric with Regex

I want to do validation for a String which can only contains alphanumeric and only one special character. I tried with (\\W).{1,1}(\\w+).
But it is true only when I start with a special character. But I can have one special character at any place in String.

Use the ^ and $ anchors to instruct the regex engine to start matching from the beginning of the string and stop matching at the end of the string, so taking your regex:
^(\\W).{1,1}(\\w+)$
Please take a look at this Oracle (Java) tutorial on regular expressions.

Try this regexp: \w*\W?\w* (Java string: "\\w*\\W?\\w*")
This expression has a drawback of matching zero-length strings. If your input must have exactly one special character, remove the question mark ? from the expression.

use matcher.find() and not matcher.match() and search for \\w and remove plus (+) because it will match all alphanumeric characters sequence in your string.If your string contains only them, your regex will match whole string.

if I understand your regex correctly, this could solve your problem:
([\w]+)([^\w])([\w]+)