iOS發送帶織物的推文，無需撰寫推文

Question

我想發推文，但我不想使用撰寫推文。

TWTRComposer *composer = [[TWTRComposer alloc] init]; 

[composer setText:@"just setting up my Fabric"]; 
[composer setImage:[UIImage imageNamed:@"fabric"]]; 

[composer showWithCompletion:^(TWTRComposerResult result) { 
    if (result == TWTRComposerResultCancelled) { 
     NSLog(@"Tweet composition cancelled"); 
    } 
    else { 
     NSLog(@"Sending Tweet!"); 
    } 
}]; 

https://dev.twitter.com/twitter-kit/ios/compose這種方法顯示一些彈出屏幕，我想送鳴叫一瞬間，我怎麼能做到這一點與面料API？這種類型是可能的？

這部分代碼不能與面料做工

ACAccountStore *accountStore = [[ACAccountStore alloc] init]; 
    ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter]; 

    [accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(BOOL granted, NSError *error) 
    { 
     if(granted) 
     { 
      NSArray *accountsArray = [accountStore accountsWithAccountType:accountType]; 

      if ([accountsArray count] > 0) 
      { 
       ACAccount *twitterAccount = [accountsArray objectAtIndex:0]; 

       TWRequest *postRequest = [[TWRequest alloc] initWithURL:[NSURL URLWithString:@"https://upload.twitter.com/1/statuses/update_with_media.json"] parameters:[NSDictionary dictionaryWithObject:self.textViewOutlet.text forKey:@"status"] requestMethod:TWRequestMethodPOST]; 

       [postRequest addMultiPartData:UIImagePNGRepresentation(image) withName:@"media" type:@"multipart/png"]; 
       [postRequest setAccount:twitterAccount]; 

       [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) 
        { 
         //show status after done 
         NSString *output = [NSString stringWithFormat:@"HTTP response status: %i", [urlResponse statusCode]]; 
         NSLog(@"Twiter post status : %@", output); 
        }]; 
      } 
     } 
    }]; 

Answer 1

https://github.com/nst/STTwitter

我用這個API，比布:)

[_twitter postStatusUpdate:@"tweet text" inReplyToStatusID:nil latitude:nil longitude:nil placeID:nil displayCoordinates:nil trimUser:nil successBlock:nil errorBlock:nil]; 

很容易發出鳴叫不作曲家可能會更容易些一個幫助別人...

Answer 2

試試這個：

func tweet(userId: String) { 
     let client = TWTRAPIClient(userID: userId) 

     let error: NSErrorPointer = NSErrorPointer() 
     let url: String = "https://api.twitter.com/1.1/statuses/update.json" 
     let message: [NSObject : AnyObject] = [ 
      "status" : "Sample Tweet Tweet!" 
     ] 

     let preparedRequest: NSURLRequest = client.URLRequestWithMethod("POST", URL: url, parameters: message, error: error) 

     client.sendTwitterRequest(preparedRequest) { (response, data, jsonError) -> Void in 
      do { 
       let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as! [String:AnyObject] 
       NSLog("%@", json) 
       print("Tweet post!") 
      } catch { 
       print("json error: \(error)") 
      } 
     } 
    } 

您必須先獲取用戶的userId。如果您使用手動按鈕註冊，請使用以下代碼：

func tweet(userId: String) { 
     let client = TWTRAPIClient(userID: userId) 

     let error: NSErrorPointer = NSErrorPointer() 
     let url: String = "https://api.twitter.com/1.1/statuses/update.json" 
     let message: [NSObject : AnyObject] = [ 
      "status" : "Sample Tweet Tweet!" 
     ] 

     let preparedRequest: NSURLRequest = client.URLRequestWithMethod("POST", URL: url, parameters: message, error: error) 

     client.sendTwitterRequest(preparedRequest) { (response, data, jsonError) -> Void in 
      do { 
       let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as! [String:AnyObject] 
       NSLog("%@", json) 
       print("Tweet post!") 
      } catch { 
       print("json error: \(error)") 
      } 
     } 
    } 

對不起，我使用Swift編寫了此代碼。