swift - UIApplicationOpenURLOptionsKey :Any return String:Any

Question

我需要“options:(NSDictionary *)options”，但这是 swift4.1 中的“options:(NSDictionary *)options”。我该怎么办？

• (BOOL)application:(UIApplication *)app openURL:(NSURL *)url options:(NSDictionary *)options NS_AVAILABLE_IOS(9_0); // no equiv. notification. return NO if the application can't open for some reason

Answer 1

目前尚不清楚您的问题是什么，但我会尝试猜测您正在寻找 open(_:options:completionHandler:) 方法。使用示例:

let url = URL(string: "https://google.com")!
UIApplication.shared.open(url) // options are equal to [:] and completionHandler is nil by default so they can be omitted

这是documentation对于这个方法。