Java 判断字符串中是否包含 emoj 表情及过滤

最近遇到一个很麻烦的问题，用户如果在客户端输入表情可能会引起一些报错，在查了一些资料后发现很多坑人的代码，我在总结了之后给出了一个比较完美的版本。

亲测可以判断绝大多数表情。

Java 实现

package com.netease.mdas.util;

import org.apache.commons.lang.StringUtils;

public class EmojiFilter {
    
	/**
	 * 判断字符串是否包含 emoji 或者 其他非文字类型的字符
	 * @param source
	 * @return
	 */
	public static boolean containsEmoji(String source) {
		int len = source.length();
		boolean isEmoji = false;
		for (int i = 0; i < len; i++) {
			char hs = source.charAt(i);
			if (0xd800 <= hs && hs <= 0xdbff) {
				if (source.length() > 1) {
					char ls = source.charAt(i + 1);
					int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;
					if (0x1d000 <= uc && uc <= 0x1f77f) {
						return true;
					}
				}
			} else {
				// non surrogate
				if (0x2100 <= hs && hs <= 0x27ff && hs != 0x263b) {
					return true;
				} else if (0x2B05 <= hs && hs <= 0x2b07) {
					return true;
				} else if (0x2934 <= hs && hs <= 0x2935) {
					return true;
				} else if (0x3297 <= hs && hs <= 0x3299) {
					return true;
				} else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c
						|| hs == 0x2b1b || hs == 0x2b50 || hs == 0x231a) {
					return true;
				}
				if (!isEmoji && source.length() > 1 && i < source.length() - 1) {
					char ls = source.charAt(i + 1);
					if (ls == 0x20e3) {
						return true;
					}
				}
			}
		}
		return isEmoji;
	}

	private static boolean isEmojiCharacter(char codePoint) {
		return (codePoint == 0x0) || (codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD)
				|| ((codePoint >= 0x20) && (codePoint <= 0xD7FF)) || ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
				|| ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF));
	}

	/**
	 * 过滤emoji 或者 其他非文字类型的字符
	 * 
	 * @param source
	 * @return
	 */
	public static String filterEmoji(String source) {
		if (StringUtils.isBlank(source)) {
			return source;
		}
		StringBuilder buf = null;
		int len = source.length();
		for (int i = 0; i < len; i++) {
			char codePoint = source.charAt(i);
			if (isEmojiCharacter(codePoint)) {
				if (buf == null) {
					buf = new StringBuilder(source.length());
				}
				buf.append(codePoint);
			}
		}
		if (buf == null) {
			return source;
		} else {
			if (buf.length() == len) {
				buf = null;
				return source;
			} else {
				return buf.toString();
			}
		}
	}
}

测试效果

public static void main(String[] args) {
	System.out.println(containsEmoji("	U+FE004	⚡")); // true
	System.out.println(containsEmoji("@netease.com")); // false
}