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

Question

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....

Answer 1

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);
            }];
        }

    }
}];