我得到一个字符串的文本大小
textSize = [[tempDict valueForKeyPath:@"caption.text"] sizeWithFont:[UIFont systemFontOfSize:12] constrainedToSize:CGSizeMake(280,CGFLOAT_MAX) lineBreakMode: NSLineBreakByWordWrapping];
我唯一的问题是,如果字符串只包含一个表情符号,我的应用程序崩溃.有没有一个简单的方法来检查emojis,或者我必须创建一个所有可能的emojis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]];
}
解决方法
尝试这个代码：
- (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;
}
