如何在swift中将字节数组转换为base64字符串？

Question

这是我在目标C中的示例代码

 -(NSString *)getImageString : (unsigned char *) charValue : (unsigned long) sizeOfBytes {                   

    uint8_t commandbyte[]={ };          

    uint8_t _allBytes[(sizeOfBytes + sizeof(commandbyte))];
    memcpy(_allBytes, charValue, sizeOfBytes);

    NSMutableData *ImageData = [[NSMutableData alloc]init];
    [ImageData appendBytes:_allBytes length:sizeof(_allBytes)];

    NSString *base64String=[self base64forData:ImageData];

    return base64String;                     
}                  

- (NSString*)base64forData:(NSData*)theData {           

    const uint8_t* input = (const uint8_t*)[theData bytes];
    NSInteger length = [theData length];

    static char table[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";

    NSMutableData* data = [NSMutableData dataWithLength:((length + 2) / 3) * 4];
    uint8_t* output = (uint8_t*)data.mutableBytes;

    NSInteger i;
    for (i=0; i < length; i += 3) {
        NSInteger value = 0;
        NSInteger j;

        for (j = i; j < (i + 3); j++) {
            value <<= 8;

            if (j < length) {
                value |= (0xFF & input[j]);
            }
        }

        NSInteger theIndex = (i / 3) * 4;
        output[theIndex + 0] = table[(value >> 18) & 0x3F];
        output[theIndex + 1] = table[(value >> 12) & 0x3F];
        output[theIndex + 2] = (i + 1) < length ? table[(value >> 6)  & 0x3F] : '=';
        output[theIndex + 3] = (i + 2) < length ? table[(value >> 0)  & 0x3F] : '=';
    }

    return [[NSString alloc] initWithData:data encoding:NSASCIIStringEncoding];
}

在这个例子中,我习惯了bytes使用sizeOfBytes然后appendBytes使用NSMutableData.

以下方法用于转换data为base64:

- (NSString*)base64forData:(NSData*)theData

它在目标C中非常简单,但是当我在swift中尝试其他指针的概念时UnsafeMutablePointer,UnsafePointer等等.

如何转换为swift 3.0 ??

你能不能请教我快速使用指针

Answer 1

您可以使用以下方法将bytes数组转换为base64String

let base64String = data!.base64EncodedString(options: NSData.Base64EncodingOptions(rawValue: 0))