java 表情符过滤

Myth丶恋晨 2022-05-15 02:18 307阅读 0赞

担心循环判断字符会影响效率，实际上使用 速度挺好。

以下为转载内容

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import org.apache.commons.lang3.StringUtils;
     
    public class EmojiFilter {
     
    	/**
    	 * 检测是否有emoji字符
    	 * 
    	 * @param source
    	 * @return 一旦含有就抛出
    	 */
    	public static boolean containsEmoji(String source) {
    		if (StringUtils.isBlank(source)) {
    			return false;
    		}
     
    		int len = source.length();
     
    		for (int i = 0; i < len; i++) {
    			char codePoint = source.charAt(i);
     
    			if (isEmojiCharacter(codePoint)) {
    				// do nothing，判断到了这里表明，确认有表情字符
    				return true;
    			}
    		}
     
    		return false;
    	}
     
    	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 (!containsEmoji(source)) {
    			return source;// 如果不包含，直接返回
    		}
    		// 到这里铁定包含
    		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);
    			} else {
    			}
    		}
     
    		if (buf == null) {
    			return source;// 如果没有找到 emoji表情，则返回源字符串
    		} else {
    			if (buf.length() == len) {// 这里的意义在于尽可能少的toString，因为会重新生成字符串
    				buf = null;
    				return source;
    			} else {
    				return buf.toString();
    			}
    		}
     
    	}
     
    }