无法通过Swifter获取Twitter用户详细信息

时间:2018-09-20 06:38:29

标签: swift4

public func authorize(withCallback callbackURL: URL,
                      presentingFrom presenting: UIViewController?,
                      forceLogin: Bool = false,
                      safariDelegate: SFSafariViewControllerDelegate? = nil,
                      success: TokenSuccessHandler?,
                      failure: FailureHandler? = nil){
     self.postOAuthRequestToken(with: callbackURL, success: { token, response in
         var requestToken = token!
         NotificationCenter.default.addObserver(forName: .swifterCallback, object: nil, queue: .main) 

{         通知中                 NotificationCenter.default.removeObserver(个体)                 presenting?.presentedViewController?.dismiss(动画:true,完成:无)                 让url = notification.userInfo![CallbackNotification.optionsURLKey]为!网址

            let parameters = url.query!.queryStringParameters
            requestToken.verifier = parameters["oauth_verifier"]

            self.postOAuthAccessToken(with: requestToken, success: { accessToken, response in
                print("accessToken:\(accessToken)------response:\(response)")
                self.client.credential = Credential(accessToken: accessToken!)
                success?(accessToken!, response)
                }, failure: failure)
        }

        let forceLogin = forceLogin ? "&force_login=true" : ""
        let query = "oauth/authorize?oauth_token=\(token!.key)\(forceLogin)"
        let queryUrl = URL(string: query, relativeTo: TwitterURL.oauth.url)!

        if #available(iOS 9.0, *) {
            if let delegate = safariDelegate ?? (presenting as? SFSafariViewControllerDelegate) {
                let safariView = SFSafariViewController(url: queryUrl)
                safariView.delegate = delegate
                safariView.modalTransitionStyle = .coverVertical
                safariView.modalPresentationStyle = .overFullScreen
                presenting?.present(safariView, animated: true, completion: nil)
            } else {
                UIApplication.shared.openURL(queryUrl)
            }
        } else {
            // Fallback on earlier versions
        }
    }, failure: failure)
}

0 个答案:

没有答案
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