Twitter SLRequest performRequestWithHandler – 无法准备URL请求

我正在尝试在我的iOS应用中实现“关注我们的Twitter”。 这是我的代码。 但它给错误“无法准备URL请求”。 请帮忙！

ACAccountStore *accountStore = [[ACAccountStore alloc] init]; ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter]; [accountStore requestAccessToAccountsWithType:accountType options:NULL completion:^(BOOL granted, NSError *error) { if (granted) { NSMutableDictionary *tempDict = [[NSMutableDictionary alloc] init]; [tempDict setValue:@"a4arpan" forKey:@"screen_name"]; [tempDict setValue:@"true" forKey:@"follow"]; SLRequest *postRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1.1/friendships/create.json"] parameters:tempDict]; ACAccount * twitterAccount = [[ACAccount alloc] initWithAccountType:accountType]; twitterAccount.username = twitterUsername; [postRequest setAccount:twitterAccount]; [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) { if (responseData) { if (urlResponse.statusCode >= 200 && urlResponse.statusCode < 300) { NSError *jsonError; NSDictionary *timelineData = [NSJSONSerialization JSONObjectWithData:responseData options:NSJSONReadingAllowFragments error:&jsonError]; if (timelineData) { NSLog(@"Timeline Response: %@\n", timelineData); } else { // Our JSON deserialization went awry NSLog(@"JSON Error: %@", [jsonError localizedDescription]); } if ([urlResponse statusCode] == 200) { UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Follow us successfull" message:nil delegate:nil cancelButtonTitle:@"Thanx" otherButtonTitles:nil, nil]; [alert show]; } else { if ([tpAppMode isEqualToString:@"sandbox"]) NSLog(@"%@", [error localizedDescription]); UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Follow us Failed" message:nil delegate:nil cancelButtonTitle:@"Thanx" otherButtonTitles:nil, nil]; [alert show]; } } else { // The server did not respond successfully... were we rate-limited? NSLog(@"The response status code is %d", urlResponse.statusCode); } } else { NSString *output = [NSString stringWithFormat:@"HTTP response status: %@ %@", [error localizedDescription], [error localizedFailureReason]]; NSLog(@"%@", output); } }]; } else { if ([tpAppMode isEqualToString:@"sandbox"]) NSLog(@"%@", [error localizedDescription]); UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Follow us Failed" message:nil delegate:nil cancelButtonTitle:@"Thanx" otherButtonTitles:nil, nil]; [alert show]; } }]; 

我已经遵循Twitter Dev所提到的所有步骤

Answer 1

我就是这样做的！

 SLRequest *request = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1.1/friendships/create.json"] parameters:[NSDictionary dictionaryWithObjects:[NSArray arrayWithObjects:@"AnatoliyGatt", @"true", nil] forKeys:[NSArray arrayWithObjects:@"screen_name", @"follow", nil]]]; [request setAccount:[[self twitterAccounts] lastObject]]; [request performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) { if(responseData) { NSDictionary *responseDictionary = [NSJSONSerialization JSONObjectWithData:responseData options:NSJSONReadingMutableContainers error:&error]; if(responseDictionary) { // Probably everything gone fine } } else { // responseDictionary is nil } }];