Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Trying to "follow" someone on Twitter using new iOS 5 API, getting 406 return error. Why?

Trying to "follow" someone on Twitter using new iOS 5 API, getting 406 return error. Why?

Is my code correct? Need to find out why this isn't working....

    - (void)followOnTwitter:(id)sender
{
    ACAccountStore *accountStore = [[ACAccountStore alloc] init];

ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];

[accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(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:@"sortitapps" forKey:@"screen_name"];
            [tempDict setValue:@"true" forKey:@"follow"];

            TWRequest *postRequest = [[TWRequest alloc] initWithURL:[NSURL URLWithString:@"http://api.twitter.com/1/friendships/create.format"] 
                                                         parameters:tempDict 
                                                      requestMethod:TWRequestMethodPOST];


            [postRequest setAccount:twitterAccount];

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

                }];
            }
        }
    }];
}

All of the code looks correct. Are the parameters incorrect? Is the URL correct? Need some direction here....

like image 730
Ethan Allen Avatar asked Nov 10 '11 20:11

Ethan Allen


4 Answers

Found the answer to my own question... I changed the URL to https://api.twitter.com/1/friendships/create.json and it worked.

Don't forget it's https, not just http.

like image 153
Ethan Allen Avatar answered Oct 23 '22 15:10

Ethan Allen


For iOS 6 twitter follow

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);
            }];
        }

    }
}];
like image 21
Amit Avatar answered Oct 23 '22 15:10

Amit


As twitter has updated their API to v1.1, the request should be made now using the URL below. Note that request should be with 'https://' and due to API v1.1 replace your 1 to 1.1 in your URL.

https://api.twitter.com/1.1/friendships/create.json

like image 42
Mohd Asim Avatar answered Oct 23 '22 16:10

Mohd Asim


you can use this code

- (BOOL)openTwitterClientForUserName:(NSString*)userName {
NSArray *urls = [NSArray arrayWithObjects:
                 @"twitter://user?screen_name={username}", // Twitter
                 @"tweetbot:///user_profile/{username}", // TweetBot
                 @"echofon:///user_timeline?{username}", // Echofon              
                 @"twit:///user?screen_name={username}", // Twittelator Pro
                 @"x-seesmic://twitter_profile?twitter_screen_name={username}", // Seesmic
                 @"x-birdfeed://user?screen_name={username}", // Birdfeed
                 @"tweetings:///user?screen_name={username}", // Tweetings
                 @"simplytweet:?link=http://twitter.com/{username}", // SimplyTweet
                 @"icebird://user?screen_name={username}", // IceBird
                 @"fluttr://user/{username}", // Fluttr
                 /** uncomment if you don't have a special handling for no registered twitter clients */
                 //@"http://twitter.com/{username}", // Web fallback, 
                 nil];

UIApplication *application = [UIApplication sharedApplication];
for (NSString *urlString in urls) {
    NSString *candidate = [urlString stringByReplacingOccurrencesOfString:@"{username}" withString:userName];
    NSURL *url = [NSURL URLWithString:candidate];
    if ([application canOpenURL:url]) {
        [application openURL:url];
        return YES;
    }
}
return NO;
}

https://gist.github.com/ChrisRicca/9144169

like image 34
SyraKozZ Avatar answered Oct 23 '22 16:10

SyraKozZ