我一直在尝试为API请求生成身份验证,如下所示:base64(sha256(payload + secret))
我一直在使用以下代码,但是生成的base64字符串不正确。
- (void)viewDidLoad
{
[super viewDidLoad];
NSString *data = @"{"
@"\"testing\":{"
@"\"uri\":\"https://example.com/something.php\","
@"\"id\":\"0\""
@"}"
@"}";
NSString *key = @"secret";
NSString *hashString = [NSString stringWithFormat:@"%@%@",data,key];
const char *cKey = [hashString cStringUsingEncoding:NSUTF8StringEncoding];
NSData *sdata = [NSData dataWithBytes:cKey length:hashString.length];
unsigned char sHMAC[64];
CC_SHA256((__bridge const void *)(sdata), sdata.length, sHMAC);
NSData *hash = [[NSData alloc] initWithBytes:sHMAC length:sizeof(sHMAC)];
NSString *s = [self base64forData:hash];
NSLog(@"Authentication: %@",s);
}
//Base64 encoding
- (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];
}
我已经为此挣扎了两个星期。
多谢您的协助!
最佳答案
JSON字典中有一个逗号。也许这搞砸了吗?
另外,请注意您的hashString
如下所示:
{"testing":{"uri":"https://example.com/something.php","id":"0",}}secret
这是故意的吗?
另外,对于您的传统C代码,之前已经完成了所有操作。参见例如this answer。
关于ios - iPhone中的API请求身份验证,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/17658387/