Created
November 26, 2012 00:54
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Check if an NSString contains an emoji character
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| - (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; | |
| } |
The U+d83e u Not included
there are some new emojis added in iOS 16.Looks like this code need to be update
搜狗输入法的一些 emoji 表情，没有包含~
for 2025 codepoint (uc) should be in range 0xF000...0x1FAFF
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I found the solution, just convert the string first
unichar unicodevalue = [text characterAtIndex:0];
NSString *str = [NSString stringWithCharacters:&unicodevalue length:1];
then use this str to judge