如何在iOS中發佈數據？

我正在做一個instagram appliaction。在那我需要發佈像圖片。我無法發佈。我正在遵循instgram API。我們需要使用以下網址如何在iOS中發佈數據？

https://api.instagram.com/v1/media/{media-id}/likes

我們需要設置媒體ID喜歡

https://api.instagram.com/v1/media/649114513807113248_1032025382/likes

媒體ID = 649114513807113248_1032025382

我的代碼發佈：

NSString *urlString = [NSString   stringWithFormat:@"https://api.instagram.com/v1/media/649114513807113248_1032025382/likes"]; 

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:urlString] 
                cachePolicy:NSURLRequestUseProtocolCachePolicy 
                timeoutInterval:60.0]; 


[request setHTTPMethod:@"POST"]; 
NSURLConnection *connection= [[NSURLConnection alloc] initWithRequest:request 
                  delegate:self]; 
[connection start];

但喜歡不張貼。誰能幫我？

來源

2014-02-10 deepti

您是否實施了NSURLConnection委託？你是否收到迴應？它說什麼？ – Anil

雅我已經實現了didReceiveData（）委託，但它顯示了一些數字，沒有有效的數據 – deepti

你可以在這裏發佈輸出？ – Anil

您可以使用AFNetworking框架以更好的方式發佈數據。樣本也在github上提供。

http://afnetworking.com/

來源

2014-02-10 07:09:29

MGInstagram和DMActivityInstagram是iOS的實用程序從您的應用程序發佈圖像至Instagram的。

// Uses our instagram instance to do a request 
- (void) postMessage:(NSString *)message mediaId:(NSString *)mediaId { 
    NSString* methodName = [NSString stringWithFormat:@"media/%@/comments", mediaId]; 
    NSMutableDictionary* params = @{"text" : message}; 
    [m_instagram requestWithMethodName:methodName params:params httpMethod:@"POST" delegate:self]; 
} 

// IGRequestDelegate method that is called once we get a response 
- (void)request:(IGRequest *)request didLoad:(id)result { 
    // post message was ok 
} 
- (void)request:(IGRequest *)request didFailWithError:(NSError *)error { 
    // post message fails 
}

從https://github.com/crino/instagram-ios-sdk/issues/27

參考，如果你是共享的圖像下面鏈接社交網絡的使用也有教程的示例應用程序

share-an-image-on-instagram-in-ios/

來源

2014-02-10 07:17:08 codercat

更好地使用AFNetworking進行後，波紋管一個樣品，請看看

-(void)commandWithParams:(NSMutableDictionary*)params onCompletion:(JSONResponseBlock)completionBlock 
{ 
    NSMutableURLRequest *apiRequest = 
     [self multipartFormRequestWithMethod:@"POST" 
             path:kAPIPath 
            parameters:params 
        constructingBodyWithBlock: ^(id <AFMultipartFormData>formData) { 
         //TODO: attach file if needed 
    }]; 

    AFJSONRequestOperation* operation = [[AFJSONRequestOperation alloc] initWithRequest: apiRequest]; 
    [operation setCompletionBlockWithSuccess:^(AFHTTPRequestOperation *operation, id responseObject) { 
     //success! 
     completionBlock(responseObject); 
    } failure:^(AFHTTPRequestOperation *operation, NSError *error) { 
     //failure :(
     completionBlock([NSDictionary dictionaryWithObject:[error localizedDescription] forKey:@"error"]); 
    }]; 

    [operation start]; 
}

來源

2014-02-10 08:33:51

如何在iOS中發佈數據？

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