3

我正在尝试执行反向 oauth 以获取服务器的 twitter 访问令牌。

我已经弄清楚如何提交请求并接收响应，但是当我这样做时，它给了我这个错误：

Error: The operation couldn’t be completed. (NSURLErrorDomain error -1012.)

我查了一下，它说这意味着用户已经取消了请求。我不确定这是怎么可能的，我不知道如何解决它。

这是我的代码：

NSTimeInterval timeStamp = [[NSDate date] timeIntervalSince1970];
    NSNumber *timeStampObj = [NSNumber numberWithDouble: timeStamp];

    NSString *oauth_nonce = [self genRandStringLength:32];
    NSString *oauth_timestamp = [timeStampObj stringValue];

    NSURL *feedURL = [NSURL URLWithString:@"https://api.twitter.com/oauth/request_token"];

    NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys: @"my key here", @"oauth_consumer_key", oauth_nonce, @"oauth_nonce", @"HMAC-SHA1", @"oauth_signature_method", oauth_timestamp, @"oauth_timestamp", @"1.0", @"oauth_version", @"reverse_auth", @"x_auth_mode", nil];

    SLRequest *twitterFeed = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:feedURL parameters:parameters];

    twitterFeed.account = self.userAccount;


    // Making the request

    [twitterFeed performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
        dispatch_async(dispatch_get_main_queue(), ^{

            // Check if we reached the reate limit

            if ([urlResponse statusCode] == 429) {
                NSLog(@"Rate limit reached");
                return;
            }

            // Check if there was an error

            if (error) {
                NSLog(@"The Error is: %@", error.localizedDescription);
                return;
            }

            // Check if there is some response data
            if (responseData) {

                NSLog(@"%@", responseData);

            }
        });
    }];

我一定缺少一些简单的东西，这使我无法完成一个项目。任何帮助都会很棒，谢谢！

4

2 回答 2

1

错误代码 -1012 可能是由于身份验证质询。就我而言，设置中存在一个 Twitter 帐户，但由于某种原因没有登录。输入帐户密码后，一切正常。

于 2014-09-10T08:33:47.117 回答
0

当我向https://api.twitter.com/oauth/request_token发送带有额外随机数和标头中签名的请求时，我遇到了这个问题。具体来说，下面的代码给了我一个 1012，但下一段代码成功了。此代码改编自Sean Cook 的反向 Twitter 身份验证示例。

/**
 *  The first stage of Reverse Auth.
 *
 *  In this step, we sign and send a request to Twitter to obtain an
 *  Authorization: header which we will use in Step 2.
 *
 *  @param completion   The block to call when finished. Can be called on any thread.
 */
- (void)_step1WithCompletion:(TWAPIHandler)completion
{
    NSURL *url = [NSURL URLWithString:TW_OAUTH_URL_REQUEST_TOKEN];
    NSDictionary *dict = @{TW_X_AUTH_MODE_KEY: TW_X_AUTH_MODE_REVERSE_AUTH,
                           TW_OAUTH_NONCE:[self nonce],
                           TW_SIGNATURE_METHOD: TW_SIGNATURE_METHOD_VALUE,
                           };

    TWSignedRequest *step1Request = [[TWSignedRequest alloc] initWithURL:url parameters:dict requestMethod:TWSignedRequestMethodPOST];

    TWDLog(@"Step 1: Sending a request to %@\nparameters %@\n", url, dict);

    [step1Request performRequestWithHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
        dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
            completion(data, error);
        });
    }

以下作品。注意中的变化dict。

/**
 *  The first stage of Reverse Auth.
 *
 *  In this step, we sign and send a request to Twitter to obtain an
 *  Authorization: header which we will use in Step 2.
 *
 *  @param completion   The block to call when finished. Can be called on any thread.
 */
- (void)_step1WithCompletion:(TWAPIHandler)completion
{
    NSURL *url = [NSURL URLWithString:TW_OAUTH_URL_REQUEST_TOKEN];
    NSDictionary *dict = @{TW_X_AUTH_MODE_KEY: TW_X_AUTH_MODE_REVERSE_AUTH};

    TWSignedRequest *step1Request = [[TWSignedRequest alloc] initWithURL:url parameters:dict requestMethod:TWSignedRequestMethodPOST];

    TWDLog(@"Step 1: Sending a request to %@\nparameters %@\n", url, dict);

    [step1Request performRequestWithHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
        dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
            completion(data, error);
        });
    }
于 2013-08-27T20:18:12.433 回答