最近遇到一個很蛋疼的問題，使用者如果在客戶端輸入表情可能會引起一些報錯，在查了一些資料後發現很多坑人的程式碼，我在總結了之後給出了一個比較完美的版本。
親測可以判斷絕大多數表情。

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));
    }

    /**
     * 過濾emoji 或者 其他非文字型別的字元
     * 
     * @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();
            }
        }
    }
}

附上測試程式碼

public class Test {

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

}