iOS - 无法使用 Twitter/Fabric 新 SDK 上传媒体

Question

我想通过我的 iOS 应用将照片发布到 Twitter。我可以在没有媒体的情况下发布推文，但是当我尝试附加媒体时会引发错误。

我正在关注 Twitter 文档，根据该文档，我首先需要将媒体上传到 https://upload.twitter.com/1.1/media/upload.json，然后我才能使用 media-id 将其与推文一起附加。

这是我上传媒体的代码。

应用在 URLRequestWithMedthod 调用时崩溃。

帮我解决问题。

UIImage *image = [UIImage imageNamed:@"shareit.png"];
NSData *imageData = UIImageJPEGRepresentation(image, 0.7);
NSString *statusesShowEndpoint = @"https://upload.twitter.com/1.1/media/upload.json";
NSDictionary *params = @{@"media" : imageData};
NSError *clientError;
NSURLRequest *request = [[[Twitter sharedInstance] APIClient] 
                         URLRequestWithMethod:@"POST"
                         URL:statusesShowEndpoint
                         parameters:params
                         error:&clientError];

if (request) {
    [[[Twitter sharedInstance] APIClient]
     sendTwitterRequest:request
     completion:^(NSURLResponse *response,
                  NSData *data,
                  NSError *connectionError) {
         if (data) {
             // handle the response data e.g.
             NSError *jsonError;
             NSDictionary *json = [NSJSONSerialization
                                   JSONObjectWithData:data
                                   options:0
                                   error:&jsonError];
             NSLog(@"%@",json);
         }
         else {
             NSLog(@"Error: %@", connectionError);
         }
     }];
}
else {
    NSLog(@"Error: %@", clientError);
}

Answer 1

嗯，这很简单。缺少的只是将图像数据转换为 base64EncodedString。这是解决方案。

   NSString *media = @"https://upload.twitter.com/1.1/media/upload.json";

   NSData *imageData = UIImageJPEGRepresentation(image, 0.9);

   NSString *imageString = [imageData base64EncodedStringWithOptions:0];
   NSError *error;
   NSURLRequest *request = [[[Twitter sharedInstance] APIClient] URLRequestWithMethod:@"POST" URL:media parameters:@{@"media":imageString} error:&error];

   [[[Twitter sharedInstance] APIClient] sendTwitterRequest:request completion:^(NSURLResponse *urlResponse, NSData *data, NSError *connectionError) {

       NSError *jsonError;
       NSDictionary *json = [NSJSONSerialization
                              JSONObjectWithData:data
                              options:0
                              error:&jsonError];
       NSLog(@"Media ID :  %@",[json objectForKey:@"media_id_string"]);

      // Post tweet With media_id
    }];

Answer 2

完成萨尼的回答

// Post tweet With media_id

mediaID = [json objectForKey:@"media_id_string"];
client = [[Twitter sharedInstance] APIClient];
message = @{@"status": title, @"wrap_links": @"true", @"media_ids": mediaID};

NSURLRequest *request = [client URLRequestWithMethod:@"POST" URL:@"https://api.twitter.com/1.1/statuses/update.json" parameters:message error:&error];

[client sendTwitterRequest:request completion:^(NSURLResponse * _Nullable response, NSData * _Nullable data, NSError * _Nullable connectionError) {

    if (connectionError) {
        NSLog(@"error %@",[connectionError localizedDescription]);
    } else {
        NSError *jsonError;
        NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:0 error:&jsonError];
        NSLog(@"json Finish! %@",json);
    }
}];

Answer 3

如果有人仍在寻求帮助，请分享 swift 2.2 (Xcoe 7.3.2) 代码。

 let client = TWTRAPIClient.init(userID:<Twitter_User_Id_Here>)
let imgData =  UIImageJPEGRepresentation(UIImage(named:"Home")!, 0.7)
    client.uploadMedia(imgData!, contentType: "image/jpeg", completion: {(media_ids,error) in

        if error == nil{
            //prints the media_ids in String. 
      //Pass them into the params of status update params ["media_ids":<MEDIA_ID>]
            print(media_ids)
        }
        else{
            print(error?.localizedDescription)
        }
    })

发送状态更新。

   let request = client.URLRequestWithMethod("POST", URL: "https://api.twitter.com/1.1/statuses/update.json", parameters: ["status":"Hello World","media_ids":<Media_Ids_Here>], error: nil)

    client.sendTwitterRequest(request, completion: {
        (response,data,error)in
        if error == nil {
            print(data)

        }
        else{
            print("Error")
        }
    })
}

Answer 4

Swift 4 最新

安装 pod 文件 pod 'TwitterKit'

import TwitterKit

https://developer.twitter.com/en/docs/media/upload-media/api-reference/post-media-upload-init.html

func PostTweetToTwitter() {
        let twitter_USERID = UserDefaults.standard.string(forKey: Constants.Twitter.TWITTER_USER_ID)

        let url = URL(string: "http://www.tinyeyeimage.com/picture/Photos/149802345.png")
        let tweetImage = try? Data(contentsOf: url!)

        let tweetString = "Welcome to the Twitter world!!"

        let uploadUrl = "https://upload.twitter.com/1.1/media/upload.json"
        let updateUrl = "https://api.twitter.com/1.1/statuses/update.json"

        let imageString = tweetImage?.base64EncodedString(options: NSData.Base64EncodingOptions())


        let client = TWTRAPIClient.init(userID: twitter_USERID)

        let requestUploadUrl = client.urlRequest(withMethod: "POST", urlString: uploadUrl, parameters: ["media": imageString], error: nil)

        client.sendTwitterRequest(requestUploadUrl) { (urlResponse, data, connectionError) -> Void in
            if connectionError == nil {
                if let mediaDict = self.dataToJSON(data: (data! as NSData) as Data) as? [String : Any] {
                    let media_id = mediaDict["media_id_string"] as! String
                    let message = ["status": tweetString, "media_ids": media_id]

                    let requestUpdateUrl = client.urlRequest(withMethod: "POST", urlString: updateUrl, parameters: message, error: nil)

                    client.sendTwitterRequest(requestUpdateUrl, completion: { (urlResponse, data, connectionError) -> Void in
                        if connectionError == nil {
                            if let _ = self.dataToJSON(data: (data! as NSData) as Data) as? [String : Any] {
                                print("Upload suceess to Twitter")
                            }
                        }

                    })
                }
            }
        }
    }