求助 js 过滤 emoji怎么实现？

Question

大神求帮助啊 一天了都

Answer 1

http://segmentfault.com/q/1010000002691280  看看是不是你要的

Answer 2

楼上 正解，摸清emoji结构，写个正则即可

Answer 3

主要是搞不清输入法里面的emoji表情是什么编码格式

Answer 4

找了好多 都不管用 用正则replace替换，怎么写都不管用 

Answer 5

- (BOOL)stringContainsEmoji:(NSString *)string
{
    __block BOOL returnValue = NO;
    [string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                               options:NSStringEnumerationByComposedCharacterSequences
                            usingBlock:^(NSString * _Nullable substring, NSRange substringRange, NSRange enclosingRange, BOOL * _Nonnull stop) {
         const unichar hs = [substring characterAtIndex:0];
         // surrogate pair
         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 <= 0x1f918)
                 {
                     returnValue = YES;
                 }
             }
         }
         else if (substring.length > 1)
         {
             const unichar ls = [substring characterAtIndex:1];
             if (ls == 0x20e3 || ls == 0xFE0F || ls == 0xd83c)
             {
                 returnValue = YES;
             }
         }
         else
         {
             // non surrogate
             if (0x2100 <= hs && hs <= 0x27ff)
             {
                 returnValue = YES;
             }
             else if (0x2B05 <= hs && hs <= 0x2b07)
             {
                 returnValue = YES;
             }
             else if (0x2934 <= hs && hs <= 0x2935)
             {
                 returnValue = YES;
             }
             else if (0x3297 <= hs && hs <= 0x3299)
             {
                 returnValue = YES;
             }
             else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50)
             {
                 returnValue = YES;
             }
         }
     }];
    return returnValue;
}

Answer 6

http://blog.csdn.net/i_qianbo/article/details/51509158看下这个方法

Answer 7

var str = "";
                     $(result).each(function(i, o){
                     o.contentShow = o.content.replace(/\[(\d)-(\d)\]/g, function(m, p1, p2){
                     return '<span class="bq" style="background-position: -' + p2*27 + 'px -' + p1*27 + 'px;"></span>';
                     });
                     });