Java正则匹配含数字、指定特殊字符及X/x的手机号

Hey there! That "Dangling meta character '*'" error happens because you're using * as a literal character but treating it like a regex quantifier (which expects to follow a pattern element). Let's build the correct regex for your needs step by step.

The Correct Java Regex #

To match phone numbers containing only your allowed characters (digits 0-9, #, *, -, _, space, (, ), +, ?, and x/X), here's the right pattern:

String phoneRegex = "^[0-9#*\\-_ ()+?xX]+$";

Breakdown of the Regex #

Let's break down each part so you understand what's happening:

• ^ : Anchors the match to the start of the string (ensures no invalid characters before your allowed ones)
• [0-9#*\\-_ ()+?xX] : A character class that matches any single allowed character:
  • 0-9: Matches any digit from 0 to 9
  • #*: Matches literal # and * (these don't need escaping inside a character class)
  • \\-: Escaped hyphen - (we escape it here to avoid confusion with a range operator; alternatively, you could place it at the start/end of the character class like [0-9#*_ ()+?xX-] to skip escaping)
  • : Matches a literal space
  • (): Matches literal parentheses (no escaping needed inside a character class)
  • +?: Matches literal + and ? (again, safe inside the character class)
  • xX: Matches both lowercase and uppercase x
• +: Requires the character class to match 1 or more times (use * instead if you want to allow empty strings, though that probably doesn't make sense for phone numbers)
• $ : Anchors the match to the end of the string (ensures no invalid characters after your allowed ones)

Example Java Code #

Here's a quick test to verify the regex works as expected:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class PhoneValidator {
    public static void main(String[] args) {
        String regex = "^[0-9#*\\-_ ()+?xX]+$";
        Pattern pattern = Pattern.compile(regex);
        
        // Test cases
        String[] testNumbers = {
            "123-456-7890",       // Valid
            "(123) 456*7890x",    // Valid
            "123+456?7890X",      // Valid
            "123_456#7890",       // Valid
            "*123-456*",          // Valid
            "123abc456",          // Invalid (contains 'abc')
            "123@456",            // Invalid (contains '@')
            ""                    // Invalid (empty string, use * instead of + to allow this)
        };
        
        for (String number : testNumbers) {
            Matcher matcher = pattern.matcher(number);
            System.out.printf("Number: '%s' -> %s%n", number, matcher.matches() ? "VALID" : "INVALID");
        }
    }
}

Why You Got the "Dangling meta character '*'" Error #

That error pops up when you use * as a quantifier without a preceding pattern element (like writing ^*abc instead of .*abc), or if you forget to put a literal * inside a character class/escape it. In your case, you likely had * outside the character class where it was being treated as a quantifier instead of a literal character.

内容的提问来源于stack exchange，提问作者Snehal Gupta