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我从Twitter的登录使用该官员Guide在TwitterKit 3.0.3授权失败
我的步骤:在应用程序管理
回调URL离开领域的空白
安装荚
pod 'Fabric'
,pod 'TwitterKit'
- Add on 的Info.plist
<key>CFBundleURLTypes</key> <array> <dict> <key>CFBundleURLSchemes</key> <array> <string>twitterkit-xxxxxxxx</string> <string>fbxxxxxxxx</string> </array> </dict> </array> <key>FacebookAppID</key> <string>xxxxx</string> <key>FacebookDisplayName</key> <string>Name</string> <key>LSApplicationQueriesSchemes</key> <array> <string>twitter</string> <string>twitterauth</string> <string>fbapi</string> <string>fb-messenger-api</string> <string>fbauth2</string> <string>fbshareextension</string> </array> <key>Fabric</key> <dict> <key>APIKey</key> <string>xxxxxxxxxx</string> <key>Kits</key> <array> <dict> <key>KitInfo</key> <dict/> <key>KitName</key> <string>Crashlytics</string> </dict> </array> </dict>
添加上
class AppDelegate
import TwitterKit .... func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool { .... Twitter.sharedInstance().start(withConsumerKey:"xxxxxx", consumerSecret:"xxxxx") } func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool { return Twitter.sharedInstance().application(app, open: url, options: options) }
添加上
Header.h
#import <TwitterKit/TwitterKit.h>
- 添加TwitterLogInButton官方Guide
当我点击按钮来登录,我得到一个错误(测试是在一个模拟器的iOS 9.3进行):
-canOpenURL: failed for URL: "twitterauth://authorize?consumer_key=xxxSAMExxx&consumer_secret=yyyyyyy&oauth_callback=twitterkit-xxxSAMExxx" - error: "(null)"
[TwitterKit] did encounter error with message "Error obtaining user auth token.": Error Domain=TWTRLogInErrorDomain Code=-1 "<?xml version="1.0" encoding="UTF-8"?>
<hash>
<error>Desktop applications only support the oauth_callback value 'oob'</error>
<request>/oauth/request_token</request>
</hash>
" UserInfo={NSLocalizedDescription=<?xml version="1.0" encoding="UTF-8"?>
<hash>
<error>Desktop applications only support the oauth_callback value 'oob'</error>
<request>/oauth/request_token</request>
</hash>
}
所示任何人都可以解决此问题? –