iOS - OAuth Twitter 工程师 - 无法创建 token

Question

在我的应用程序中，我正在尝试将发布消息集成到 Twitter。(我在我的推特应用程序中生成了消费者 key 和消费者 secret 。
我正在使用以下代码。

#pragma mark SA_OAuthTwitterEngineDelegate
- (void) storeCachedTwitterOAuthData: (NSString *) data forUsername: (NSString *) username {
    NSUserDefaults          *defaults = [NSUserDefaults standardUserDefaults];

    [defaults setObject: data forKey: @"authData"];
    [defaults synchronize];
}

- (NSString *) cachedTwitterOAuthDataForUsername: (NSString *) username {
    return [[NSUserDefaults standardUserDefaults] objectForKey: @"authData"];
}

- (void) twitterOAuthConnectionFailedWithData: (NSData *) data
{
    NSLog(@"Failed");
}
//=============================================================================================================================
#pragma mark SA_OAuthTwitterControllerDelegate
- (void) OAuthTwitterController: (SA_OAuthTwitterController *) controller authenticatedWithUsername: (NSString *) username {
    NSLog(@"Authenicated for %@", username);
}

- (void) OAuthTwitterControllerFailed: (SA_OAuthTwitterController *) controller {
    NSLog(@"Authentication Failed!");
}

- (void) OAuthTwitterControllerCanceled: (SA_OAuthTwitterController *) controller {
    NSLog(@"Authentication Canceled.");
}

它授权成功，但是下面的委托(delegate)方法打印的是空用户名。

- (void) OAuthTwitterController: (SA_OAuthTwitterController *) controller authenticatedWithUsername: (NSString *) username

以下是我收到的回复:

2011-12-29 00:01:57.965 OAuthTwitterDemo[4656:207] Failed
2011-12-29 00:01:57.966 OAuthTwitterDemo[4656:207] Authenicated for (null)

有人可以告诉我我的代码中有什么错误吗？

Answer 1

尝试把这个函数中的http换成https

- (SA_OAuthTwitterEngine *) initOAuthWithDelegate: (NSObject *) delegate {
if (self = (id) [super initWithDelegate: delegate]) {
        self.requestTokenURL = [NSURL URLWithString: @"https://twitter.com/oauth/request_token"];
        self.accessTokenURL = [NSURL URLWithString: @"https://twitter.com/oauth/access_token"];
        self.authorizeURL = [NSURL URLWithString: @"https://twitter.com/oauth/authorize"];
    }
return self;
}