correct usage of String.matches() and regex

Im trying to check if my input contains anything but A-Z a-z , . ' - and whitespace.
I guess it's a simple mistake because im quite a rookie when it comes to regex.

public class Test {
  public static void main(String args[]) {
	  doesMatch(0,"Hello ', . - ");
	  doesMatch(1,"1Hello1");
	  doesMatch(2,"23123");
	  doesMatch(3,"§!$'##");
	  doesMatch(4,"pe33teramjd");
	  doesMatch(5,"3pe33teramjd");
	  doesMatch(6,"pe33teramjd3");
	  doesMatch(7,"yup py");
    }
  
  static void doesMatch(int number,String input){
	  System.out.println("Number: "+number+" | "+input.matches("[^A-Za-z,.'\\s-]"));
  }
}

output:

Number: 0 | false
Number: 1 | false
Number: 2 | false
Number: 3 | false
Number: 4 | false
Number: 5 | false
Number: 6 | false
Number: 7 | false

desired output:

Number: 0 | false
Number: 1 | true
Number: 2 | true
Number: 3 | true
Number: 4 | true
Number: 5 | true
Number: 6 | true
Number: 7 | false

Explanation

> Im trying to check if my input contains anything but A-Z a-z , . ' - and whitespace.

Or, the reverse logic: You are trying to verify whether the text only consists of A-Z a-z , . ' - and whitespace.

You likely intended to repeat your regex pattern. So [...]+ instead of just [...], which would match a single character only.

Then, get rid of the ^ which would negate the expression. Also, you have to escape the ., so \\., otherwise it matches any character and not the . (dot).

The regex pattern now is "[A-Za-z,\\.'\\s-]+". Now you receive the following output:

Number: 0 | true
Number: 1 | false
Number: 2 | false
Number: 3 | false
Number: 4 | false
Number: 5 | false
Number: 6 | false
Number: 7 | true

Just negate the boolean with ! and you are done.

For reference, the full code:

public class Test {
  public static void main(String args[]) {
      doesMatch(0,"Hello ', . - ");
      doesMatch(1,"1Hello1");
      doesMatch(2,"23123");
      doesMatch(3,"§!$'##");
      doesMatch(4,"pe33teramjd");
      doesMatch(5,"3pe33teramjd");
      doesMatch(6,"pe33teramjd3");
      doesMatch(7,"yup py");
    }
  
  static void doesMatch(int number,String input){
      System.out.println("Number: "+number+" | "+ !input.matches("[A-Za-z,\\.'\\s-]+"));
  }
}

Without negation

If you do not want to negate the final result with ! you have to correctly negate the regex pattern, which would informally be "if any character is not A-Z a-z , . ' - or whitespace then it matches"

A pattern checking that could be

".*[^A-Za-z,\\.'\\s-].*"

The .* means any character sequence.

Search instead of full match

Instead of attempting a full-match, this is likely better done by having a regex search using find():

Pattern pattern = Pattern.compile("[^A-Za-z,\\.'\\s-]");
...
boolean result = pattern.matcher(input).find();

This will now tell you whether it found anything matching the regex, so any character that is not A-Za-z,\\.'\\s-.