带标志表情符号的NSString长度 [英] NSString length with flag emojis

问题描述

我正在尝试获取NSString的长度，该长度包含一堆表情符号，包括标志字符。现在，我知道如何获取包含表情符号的字符串的长度：

I'm trying to get the length of an NSString that contains a bunch of emojis, including the flag characters. Now I know how to get the length of a string containing emojis:

__block NSInteger length = 0;
[string enumerateSubstringsInRange:range
                           options:NSStringEnumerationByComposedCharacterSequences
                        usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
    length++;
}];

但是， NSStringEnumerationByComposedCharacterSequences在这种情况下不起作用。普通的表情符号由两个字符组成。标志表情符号由两个类似表情符号的字符组成。因此，一个没有任何枚举技术的标志表情符号的长度为4。

However, "NSStringEnumerationByComposedCharacterSequences" doesn't work in this case. Normal emojis are composed of two characters put together. The flag emojis are composed of two emoji-like characters put together. Thus the length of one flag emoji without any enumeration technique is 4.

如何获取包含标志表情符号的NSString的正确长度（字符数）？

How do I get the proper length (number of characters) of an NSString containing flag emojis?

推荐答案

我尝试了以下解决方案。
对于普通表情符号，它返回2，对于标志，则返回4。
这也适用于普通字符和特殊字符。

I tried the below solution. For normal emoji it returns 2 and for flags it will give 4. Also this will work with normal chars and special chars.

const char *cString = [string UTF8String];

int textLength = (int)strlen(cString);

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