java 判断字符串中是否包含emoj表情及过滤，完美解决。

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最近遇到一个很蛋疼的问题，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。

package com.luo.dtqjh.utils;import org.apache.commons.lang.StringUtils;public class EmojiFilter {    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();            }        }    }}

附上测试代码

public class Test {

public static void main(String[] args) {    String string = "��都嗨��、齐静��给你��";    System.out.println(EmojiFilter.containsEmoji(string));    System.out.println(EmojiFilter.filterEmoji(string));}

}

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