java - 用户名的 Java 正则表达式

Answer 1

For my answer, I want to use a simpler regex similar to yours: [A-Z[^!]]+, which means "At least once: (a character from A to Z) or (a character that is not '!').
Note that "not '!'" already includes A to Z. So everything in the outer character group([A-Z...) is pointless.

Try [\p{Alpha}'-.]+ and compile the regex with the Pattern.UNICODE_CHARACTER_CLASS flag.

Answer 2

Use: (?=.*[@#$%&\s]) - Return true when atleast one special character (from set) and also if username contain space.

you can add more special character as per your requirment. For Example:

String str = "k$shor";
String regex = "(?=.*[@#$%&\\s])";
Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(str);
System.out.println(matcher.find()); => gives true