"canOpenURL" is acting weird... or am I?

Question

I find it weird that though canOpenURL(myURL) is false while open(myURL) perfectly redirected to the app I want. What am I doing wrong here?

override func viewDidAppear(_ animated: Bool) {
    let instagramUrl = URL(string: "instagram://app")!
    print(UIApplication.shared.canOpenURL(instagramUrl))    // false
    if UIApplication.shared.canOpenURL(instagramUrl) {
        UIApplication.shared.open(instagramUrl, options: [:], completionHandler: nil)
    } else {
        UIApplication.shared.open(URL(string: "http://instagram.com")!, options: [:], completionHandler: nil)    // Instagram successfully launched???
        // UIApplication.shared.open(instagramUrl, options: [:], completionHandler: nil)        // This works too
        print("Open failed.")    // Open failed.
    }
}

Console:

-canOpenURL: failed for URL: "instagram://app" - error: "This app is not allowed to query for scheme instagram"

[MC] System group container for systemgroup.com.apple.configurationprofiles path is /private/var/containers/Shared/SystemGroup/systemgroup.com.apple.configurationprofiles

[MC] Reading from public effective user settings.

My Info.plist as below

Possible duplicate of [how to open instagram app on button click](https://stackoverflow.com/questions/38803801/how-to-open-instagram-app-on-button-click) — Jack, Jul 15 '18 at 13:35

score 3 · Accepted Answer · answered Jul 15 '18 at 13:34

3

You have to include your URL scheme into LSApplicationQueriesSchemes, not the URL itself.

So, instead of all the items in your current LSApplicationQueriesSchemes, you just need one item with the value instagram and it will work.

answered Jul 15 '18 at 13:34

Papershine

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Thank you so much!!! You literally saved my night!!! I am using the scheme stuff for the first time so I was somewhat lost. Thank you! – Lambdalex Jul 15 '18 at 13:46

"canOpenURL" is acting weird... or am I?

1 Answers1