我正在开发一个需要获取 youtube channel 并订阅它的 iOS 应用程序。 我可以获得没有 OAuth2 的 channel 列表,但要订阅它,我需要通过 OAuth2 标准获取访问 token 。
我可以通过一些手工获得访问 token 。但是,当我将此 token 发送到 header 并请求正文进行订阅时,它会给我“错误请求(400)”错误。
NSString *URLString = [NSString stringWithFormat"https://www.googleapis.com/youtube/v3/subscriptions?part=snippet&key=%@", @"[mykey]"];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
AFHTTPRequestSerializer *requestSerializer = [AFHTTPRequestSerializer serializer];
[requestSerializer setValue:[NSString stringWithFormat"Bearer %@",[self accessToken]] forHTTPHeaderField"Authorization"];
[requestSerializer setValue"application/json" forHTTPHeaderField"Content-Type"];
manager.requestSerializer = requestSerializer;
NSDictionary *parameters = @{@"snippet"{
@"resourceId"{
@"channelId":[[[videoCountDetailArray objectAtIndex:0] valueForKey"snippet"] valueForKey"channelId"],
@"kind""youtube#channel"
}
}
};
[manager POST:URLString parameters:parameters success:^(AFHTTPRequestOperation *operation, id responseObject) {
NSLog(@"response ; %@",responseObject);
} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(@"failed");
}];
}
Error Domain=com.alamofire.error.serialization.response Code=-1011 "Request failed: bad request (400)" UserInfo={com.alamofire.serialization.response.error.response=<NSHTTPURLResponse: 0x7fcfbccae3a0> { URL: https://www.googleapis.com/youtube/v3/subscriptions?part=snippet&key=myKey } { status code: 400, headers {
"Cache-Control" = "private, max-age=0";
"Content-Encoding" = gzip;
"Content-Length" = 118;
"Content-Type" = "application/json; charset=UTF-8";
Date = "Wed, 25 May 2016 09:28:57 GMT";
Expires = "Wed, 25 May 2016 09:28:57 GMT";
Server = GSE;
Vary = "Origin, X-Origin";
"alt-svc" = "quic=\":443\"; ma=2592000; v=\"34,33,32,31,30,29,28,27,26,25\"";
"alternate-protocol" = "443:quic";
"x-content-type-options" = nosniff;
"x-frame-options" = SAMEORIGIN;
"x-xss-protection" = "1; mode=block";
} },
如有任何帮助,将不胜感激。
我发现了问题。有两个错误。
其实我是在设置不同的requestSerializer,传递不同的序列化器。
我正在将 APIKey 传递给 url 进行订阅,但不需要传递它。
我的新代码是
NSString *URLString = [NSString stringWithFormat:@"https://www.googleapis.com/youtube/v3/subscriptions?part=snippet"];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
AFHTTPRequestSerializer *requestSerializer = [AFJSONRequestSerializer serializer];
[requestSerializer setValue:[NSString stringWithFormat:@"Bearer %@",[self accessToken]] forHTTPHeaderField:@"Authorization"];
[requestSerializer setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
manager.requestSerializer = requestSerializer;
NSDictionary *parameters = @{@"snippet":@{
@"resourceId":@{
@"channelId":[[[videoCountDetailArray objectAtIndex:0] valueForKey:@"snippet"] valueForKey:@"channelId"],
@"kind":@"youtube#channel"
}
}
};
NSLog(@"parameters : %@",parameters);
[manager POST:URLString parameters:parameters success:^(AFHTTPRequestOperation *operation, id responseObject) {
NSLog(@"response ; %@",responseObject);
} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(@"failed");
}];
关于ios - Youtube API iOS : Request failed: bad request (400),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37433203/
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