UIApplicationOpenURLOptionsKey:Any return String:Any

0
votes

I need "options:(NSDictionary *)options" ,but this is "options:(NSDictionary *)options" in swift4.1. What should I do?

  • (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
1
swift
The code is Objective-C. Please read the Swift documentation – vadian

1 Answers

0
votes

It's not clear what's your question about, but I'll try to guess that you're looking for open(_:options:completionHandler:) method. Usage example:

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

Here's documentation for this method.