項目中原來使用的方法：

- (BOOL)ContainsEmoji:(NSString *)inputString 
{
    __block BOOL isEomji = NO;
    [inputString enumerateSubstringsInRange:NSMakeRange(0, [inputString length]) options:NSStringEnumerationByComposedCharacterSequences usingBlock:
     ^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
        const unichar hs = [substring characterAtIndex:0];
        if (0xd800 <= hs && hs <= 0xdbff) {
            if (substring.length > 1) {
                const unichar ls = [substring characterAtIndex:1];
                const int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;
                if (0x1d000 <= uc && uc <= 0x1F77F) {
                    isEomji = YES;
                }
            }
        } else if (substring.length > 1) {
            const unichar ls = [substring characterAtIndex:1];
            if (ls == 0x20e3 || ls == 0xfe0f) {
                isEomji = YES;
            }
        } else {
            if (0x2100 <= hs && hs <= 0x27ff) {
                isEomji = YES;
            } else if (0x2B05 <= hs && hs <= 0x2b07) {
                isEomji = YES;
            } else if (0x2934 <= hs && hs <= 0x2935) {
                isEomji = YES;
            } else if (0x3297 <= hs && hs <= 0x3299) {
                isEomji = YES;
            } else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50|| hs == 0x231a ) {
                isEomji = YES;
            }
        }
    }];
    return isEomji;
}

發(fā)現(xiàn)有一些emoji會遺漏過去拴曲，如：?? ?? ?? ?? ?? ?? 等苔巨；

調(diào)試發(fā)現(xiàn)，被識別出來的在 0x1d000 <= uc && uc <= 0x1F77F 這兩個值之間，而?? 的值更大遭贸。于是找到 UNICODE EMOJI，知道最新版本的emoji是14.0遣妥，發(fā)布日期：2021-08-25品姓。

然后在文檔中找到14.0版本中emoji編碼的最大值：0x1FAF6，于是將識別的范圍改為：0x1d000 <= uc && uc <= 0x1FAF6本缠，測試過后斥扛，上述遺漏的emoji都被攔截了。

麻煩的點是emoji更新后丹锹，需要手動更新代碼稀颁，有不對、遺漏的地方或者大家有更好的方法楣黍，希望不吝賜教匾灶。

----------------------------------------------------------------這是一條分割線----------------------------------------------------------
發(fā)現(xiàn)一個大佬用正則表達式寫的簡潔方法：

NSString *inputString = @"abcd123?? ?? ?? ?? ?? ??";
NSRegularExpression *regularExpression = [NSRegularExpression regularExpressionWithPattern:@"[^\\u0020-\\u007e\\u00a0-\\u00be\\u2e80-\\ua4cf\\uf900-\\ufaff\\ufe30-\\ufe4f\\uff00-\\uffef\\u0080-\\u009f\\u2000-\\u201f\r\n]" options:0 error:nil];
NSString *result = [regularExpression stringByReplacingMatchesInString:inputString options:0 range:NSMakeRange(0, inputString.length) withTemplate:@""];

另外可以直接不讓emoji輸入，不過 currentInputMode 官方不建議使用了：

-(BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string
{
    if ([[[UITextInputMode currentInputMode] primaryLanguage] isEqualToString:@"emoji"]) {
        return false;
    }
    return true;
}