是否可以在 ios 中使用 SLRequest 在 Facebook 上发出好友请求？

Question

我正在使用SLRequest 获取用户详细信息并在 facebook 和 twitter 上分享。我还使用 Linkedin SDK 和 Google+ SDK 来获取各自社交网络上的用户详细信息。我的问题是

1. 是否可以在 facebook 上使用SLRequest 发送好友请求？
2. 是否可以使用 SLRequest 在 Twitter 上关注某人？
3. 是否可以使用 SDK 在linkedin 上连接一个人？
4. 是否可以使用 SDK 在 google+ 上添加人员？

如果可能，请给我一个方法来做到这一点。谢谢。

Answer 1

找了半天，找到了我的问题的解决办法，

1. Facebook graph api 2.0版本不支持好友请求（参考：https://developers.facebook.com/docs/dialogs/friends/v2.0）。

2. 我们可以用 twitter 来做如下

   ACAccountStore *accountStore = [[ACAccountStore alloc] init];
   
   ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];
   
   [accountStore requestAccessToAccountsWithType:accountType options:nil completion:^(BOOL granted, NSError *error) {
   if(granted) {
       // Get the list of Twitter accounts.
       NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];
   
       // For the sake of brevity, we'll assume there is only one Twitter account present.
       // You would ideally ask the user which account they want to tweet from, if there is more than one Twitter account present.
       if ([accountsArray count] > 0) {
           // Grab the initial Twitter account to tweet from.
           ACAccount *twitterAccount = [accountsArray objectAtIndex:0];
   
           NSMutableDictionary *tempDict = [[NSMutableDictionary alloc] init];
           [tempDict setValue:@"MohammadMasudRa" forKey:@"screen_name"];
           [tempDict setValue:@"true" forKey:@"follow"];
           NSLog(@"*******tempDict %@*******",tempDict);
   
           //requestForServiceType
   
           SLRequest *postRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1/friendships/create.json"] parameters:tempDict];
           [postRequest setAccount:twitterAccount];
           [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
               NSString *output = [NSString stringWithFormat:@"HTTP response status: %i Error %d", [urlResponse statusCode],error.code];
               NSLog(@"%@error %@", output,error.description);
           }];
       }
   
   }
   

   }];

3. 我们可以使用 Linkedin 来做到这一点。

   NSURL *url = [NSURL URLWithString:@"http://api.linkedin.com/v1/people/~/mailbox"];
   
   OAMutableURLRequest *request =
   [[OAMutableURLRequest alloc] initWithURL:url
                               consumer:oAuthLoginView.consumer
                                  token:oAuthLoginView.accessToken
                               callback:nil
                      signatureProvider:nil];
   [request setHTTPMethod:@"POST"];
   NSString *messageToPerson = @"/people/email=test123@test.com";
   NSDictionary *person = [[NSDictionary alloc] initWithObjectsAndKeys:[[NSDictionary alloc] initWithObjectsAndKeys:messageToPerson,@"_path",@"test123",@"first-name",@"test",@"last-name",nil], @"person",nil];
   NSArray *valueArray = [[NSArray alloc] initWithObjects:person,nil];
   NSDictionary *values = [[NSDictionary alloc] initWithObjectsAndKeys:valueArray,@"values", nil];
   NSDictionary *ir = [[NSDictionary alloc] initWithObjectsAndKeys:[[NSDictionary alloc] initWithObjectsAndKeys:@"friend",@"connect-type",nil], @"invitation-request",nil];
   NSDictionary *update = [[NSDictionary alloc] initWithObjectsAndKeys:values,@"recipients",@"Invitation",@"subject",@"ConnectWithMe",@"body", ir, @"item-content", nil];
   [request setValue:@"json" forHTTPHeaderField:@"x-li-format"];
   [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
   
   NSString *updateString = [update JSONString];
   
   [request prepare];
   
   [request setHTTPBodyWithString:updateString];
   
   OADataFetcher *fetcher = [[OADataFetcher alloc] init];
   
   [fetcher fetchDataWithRequest:request
                    delegate:self
           didFinishSelector:@selector(postUpdateApiCallResult1:didFinish:)
             didFailSelector:@selector(postUpdateApiCallResult1:didFail:) withPrepare:NO];
   

4. 我无法通过 google plus add to circle 获取详细信息。