iOS Google登录错误

New*_*ter 10 ios google-signin

我正在尝试实现谷歌登录按钮,当我将以下行添加到我的AppDelegate.swift文件时,我收到此错误:

无法使用索引类型为"String"的类型'[String:AnyObject]'下标值

任何想法这个代码有什么问题?顺便说一句,这段代码只是从谷歌页面复制并粘贴:https: //developers.google.com/identity/sign-in/ios/sign-in?configurationdéver= swtift

func application(application: UIApplication,
    openURL url: NSURL, options options: [String: AnyObject]) -> Bool {
        return GIDSignIn.sharedInstance().handleURL(url,
            sourceApplication: options[UIApplicationOpenURLOptionsSourceApplicationKey],
            annotation: options[UIApplicationOpenURLOptionsAnnotationKey])
}
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iOS Google登录错误

谢谢

lon*_*tle 12

handleURL正在为sourceApplication和annotation查找类型为"String"的参数,但options提供了"AnyObject".将这些字典值转换为"String"应该这样做.

试试这个:

更新:Swift 3

func application(_ application: UIApplication,
                 open url: URL, options: [UIApplicationOpenURLOptionsKey: Any]) -> Bool
   return GIDSignIn.sharedInstance().handle(url,
       sourceApplication: options[UIApplicationOpenURLOptionsKey.sourceApplication] as? String,
       annotation: options[UIApplicationOpenURLOptionsKey.annotation] as? String)
}
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斯威夫特2

func application(application: UIApplication,
    openURL url: NSURL, options: [String: AnyObject]) -> Bool {
        return GIDSignIn.sharedInstance().handleURL(url,
            sourceApplication: options[UIApplicationOpenURLOptionsSourceApplicationKey] as? String,
            annotation: options[UIApplicationOpenURLOptionsAnnotationKey] as? String)
}
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这也让我永远.希望有所帮助!


小智 8

Xcode 8 Swift 3

func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool {
  return GIDSignIn.sharedInstance().handle(
  url,
  sourceApplication: options[UIApplicationOpenURLOptionsKey.sourceApplication] as! String,
  annotation: options[UIApplicationOpenURLOptionsKey.annotation])
}
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