How to open URL in android?

[Arnav-arw]

As the title suggests,

 #if SKIP
              
  #else
  let url = URL(string: zen.articleUrl)!
  UIApplication.shared.open(url)
  #endif

[marcprux]

If you are in a SwiftUI view, the recommended way to open a URL is like:

public struct ContentView: View {
    @Environment(\.openURL) var openURLAction

    public var body: some View {
        Button("Launch URL") {
            let url = URL(string: "https://skip.tools")!
            openURLAction(url)
        }
    }
}

If you need to call this from a non-SwiftUI view (e.g., from model code), note that we do implement UIApplication.shared.open, but there was a bug with it that prevented it from working until the latest version of SkipUI (0.10.7). So once you update your packages, this should also work:

func openSkipURL() {
    Task {
        let url = URL(string: "https://skip.tools")!
        await UIApplication.shared.open(url)
    }
}

P.S. Note that you will need to ensure your AndroidManifest.xml file contains the android.permission.INTERNET permission:

<?xml version="1.0" encoding="utf-8"?>
<!-- This AndroidManifest.xml template was generated by Skip -->
<manifest xmlns:android="http://schemas.android.com/apk/res/android" xmlns:tools="http://schemas.android.com/tools">
    <!-- permissions needed for using the internet or an embedded WebKit browser -->
    <uses-permission android:name="android.permission.INTERNET" />
</manifest>

[Arnav-arw]

Thanks! Both works well.