检查字符串中是否包含表情符号

Question

我用这个获得字符串的文本大小

textSize = [[tempDict valueForKeyPath:@"caption.text"] sizeWithFont:[UIFont systemFontOfSize:12] constrainedToSize:CGSizeMake(280, CGFLOAT_MAX) lineBreakMode: NSLineBreakByWordWrapping];

我唯一的问题是,如果字符串只包含表情符号,我的应用程序崩溃.有没有一种简单的方法来检查表情符号,或者我是否必须创建一个包含所有可能的表情符号的数组,然后使用它来检查它们？

错误:

-[NSNull sizeWithFont:constrainedToSize:lineBreakMode:]: unrecognized selector sent to instance 0x3aa88a60


if ([tempDict valueForKeyPath:@"caption.text"]){
            NSLog(@"%@", [tempDict valueForKeyPath:@"caption"]);
            //Measure the message label box height
            textSize = [[tempDict valueForKeyPath:@"caption.text"] sizeWithFont:[UIFont systemFontOfSize:12] constrainedToSize:CGSizeMake(280, CGFLOAT_MAX) lineBreakMode: NSLineBreakByWordWrapping];
            int height = 320 + 20 + textSize.height;
            [cellHeight addObject:[NSNumber numberWithInt:height]];
}

Answer 1

试试这段代码:

- (BOOL)stringContainsEmoji:(NSString *)string {
    __block BOOL returnValue = NO;
    [string enumerateSubstringsInRange:NSMakeRange(0, [string length]) options:NSStringEnumerationByComposedCharacterSequences usingBlock:
     ^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *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 <= 0x1f77f) {
                     returnValue = YES;
                 }
             }
         } else if (substring.length > 1) {
             const unichar ls = [substring characterAtIndex:1];
             if (ls == 0x20e3) {
                 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 2

@Simha.IC答案很棒.

但它不适用于新的iOS 9.1 emojis'.

Simha.IC片段没有检测到这些:
☂️✝️✡️☯️

所以为了解决这个问题,我稍微调整了一下代码并用它创建了一个类别:

- (BOOL)emo_containsEmoji
{
    __block BOOL containsEmoji = NO;

    [self enumerateSubstringsInRange:NSMakeRange(0,
                                                 [self length])
                             options:NSStringEnumerationByComposedCharacterSequences
                          usingBlock:^(NSString *substring,
                                       NSRange substringRange,
                                       NSRange enclosingRange,
                                       BOOL *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 <= 0x1f9c0)
                 {
                     containsEmoji = YES;
                 }
             }
         }
         else if (substring.length > 1)
         {
             const unichar ls = [substring characterAtIndex:1];
             if (ls == 0x20e3 ||
                 ls == 0xfe0f ||
                 ls == 0xd83c)
             {
                 containsEmoji = YES;
             }
         }
         else
         {
             // non surrogate
             if (0x2100 <= hs &&
                 hs <= 0x27ff)
             {
                 containsEmoji = YES;
             }
             else if (0x2B05 <= hs &&
                      hs <= 0x2b07)
             {
                 containsEmoji = YES;
             }
             else if (0x2934 <= hs &&
                      hs <= 0x2935)
             {
                 containsEmoji = YES;
             }
             else if (0x3297 <= hs &&
                      hs <= 0x3299)
             {
                 containsEmoji = YES;
             }
             else if (hs == 0xa9 ||
                      hs == 0xae ||
                      hs == 0x303d ||
                      hs == 0x3030 ||
                      hs == 0x2b55 ||
                      hs == 0x2b1c ||
                      hs == 0x2b1b ||
                      hs == 0x2b50)
             {
                 containsEmoji = YES;
             }
         }
     }];

    return containsEmoji;
}

我的类别可以计算:

• 如果字符串有表情符号
• 表情符号的数量
• 每个表情符号的范围

我创建了一个以类别为例的小项目.

我还创建了一个类别的pod,也许你想尝试一下.

Answer 3

请参阅这个答案How to get NSString size when NSString include emojis?

如果您只有表情符号，那么您不应该使用sizeWithFont. 使用标签或其他东西。

NSString 不适用于表情符号，仅适用于文本。如果您有表情符号，那么您应该使用[label sizeToFit]或[label sizeThatFits:]

另外，您可以使用以下代码：

 id text = [tempDict valueForKeyPath:@"caption.text"]

 if (![text isKindOfClass:[NSNull class]]) {
 ...
 }

因为，你从字典中得到 NSNull，但它不等于 nil，并且你的条件有效。