iOS emoji表情转码 或者判断

如果项目中有评论或者信息恢复的地方，往往会用到emoji，有时候如后台不支持emoji，就会显示乱码错误，我们可以把emoji转成unicode编码或者utf8编码格式传给服务器。当然如果后台服务器接收的时候能做好判断识别最好，我们这边后台是支持的，我仅记录一下方法，以备不时之需。

先定义一个UITextView 并设置代理

设定一个宏定义，用来判断emoji

#define MULITTHREEBYTEUTF16TOUNICODE(x,y) (((((x ^ 0xD800) << 2) | ((y ^ 0xDC00) >> 8)) << 8) | ((y ^ 0xDC00) & 0xFF)) + 0x10000

下面写代理方法实现的内容

- (BOOL)textView:(UITextView *)textView shouldChangeTextInRange:(NSRange)range replacementText:(NSString *)text
{
NSString *hexstr = @"";

for (int i=0;i< [text length];i++)
{
hexstr = [hexstr stringByAppendingFormat:@"%@",[NSString stringWithFormat:@"0x%1X ",[text characterAtIndex:i]]];
}
NSLog(@"UTF16 [%@]",hexstr);

hexstr = @"";

long slen = strlen([text UTF8String]);

for (int i = 0; i < slen; i++)
{
//fffffff0 去除前面六个F & 0xFF
hexstr = [hexstr stringByAppendingFormat:@"%@",[NSString stringWithFormat:@"0x%X ",[text UTF8String][i] & 0xFF ]];
}
NSLog(@"UTF8 [%@]",hexstr);

hexstr = @"";

if ([text length] >= 2) {

for (int i = 0; i < [text length] / 2 && ([text length] % 2 == 0) ; i++)
{
// three bytes
if (([text characterAtIndex:i*2] & 0xFF00) == 0 ) {
hexstr = [hexstr stringByAppendingFormat:@"Ox%1X 0x%1X",[text characterAtIndex:i*2],[text characterAtIndex:i*2+1]];
}
else
{// four bytes
hexstr = [hexstr stringByAppendingFormat:@"U+%1X ",MULITTHREEBYTEUTF16TOUNICODE([text characterAtIndex:i*2],[text characterAtIndex:i*2+1])];
}

}
NSLog(@"(unicode) [%@]",hexstr);
}
else
{
NSLog(@"(unicode) U+%1X",[text characterAtIndex:0]);
}

return YES;
}

在输入的时候，会自动把输入内容转成相应的格式。

如果在有些地方不需要输入emoji表情，可以做相关限制。

我这边用到的是，如果用户输入emoji表情的时候，会给出提示

//是否含有表情
- (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];
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 {
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;
}

通过调用该方法，如果返回的是YES则输入内容含有emoji，反之。