terminate app after opening safari - swift

Question

I have a button in my app that opens a link via safari. There are some tasks user does in web and finally returns to app on pressing a button in browser. what I need to do is to terminate app when safari opens so that when user comes back form browser, app inits without previous data. That's how I open link:

let adrs = "mylink"
UIApplication.shared.openURL(NSURL(string: adrs) as! URL)

How can I achieve this?

can you show me your link? – Mukesh Lokare Feb 20 '17 at 14:04 — Mukesh Lokare, Feb 20 '17 at 14:04
@Mukesh That's not an issue. suppose it's http://google.com – Hos Ap Feb 20 '17 at 14:14 — Hos Ap, Feb 20 '17 at 14:14

score 1 · Accepted Answer · answered Feb 20 '17 at 13:58

1

Technically you can just exit the program (e.g., exit(0)). However, you should not do so, as iOS apps should not terminate on their own. Rather clean up your UI, e.g., remove all view controllers and create new ones.

answered Feb 20 '17 at 13:58

Arkku

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how to clean all VCs and create new ones? and can i do it after opening the link? – Hos Ap Feb 20 '17 at 14:01
You could set the `rootViewController` of the main window (save a reference to it in the App Delegate). Or set the `viewControllers` of your navigation controller, if you are using one. – Arkku Feb 20 '17 at 14:15
can i go to root after opening url? – Hos Ap Feb 20 '17 at 14:16
@HosAp Why not? – Arkku Feb 20 '17 at 14:16
I thought codes below openURL would be ignored. that works . thanks – Hos Ap Feb 20 '17 at 14:21

terminate app after opening safari - swift

1 Answers1