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

Question

This is the code I use:

let instagramURL = NSURL(string: "instagram://app")
if UIApplication.shared.canOpenURL(instagramURL! as URL) {
  //Code
} else {
  //Showing message "Please install the Instagram application"
}

I am getting unsuccessful to enter in if loop.

I get this error:

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

I have also Login with Instagram in my device.

Answer 1

Right click on your plist file and open it as source code . Then copy and paste below code:

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>instagram</string>
</array>

Note: One thing you have to keep in mind that it will not work on simulator. You need a real device for this.

Answer 2

Open your plist as source code and paste following code:

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>instagram</string>
</array>

Answer 3

The issue is that you are not registering the URL scheme in the info.plist file.

Please add this LSApplicationQueriesSchemes and add instagram in your info.plist and it will work.

Answer 4

For those who try to open the app using custom URL Scheme (assume that app FirstApp opens SecondApp):

• In FirstApp add LSApplicationQueriesSchemes with URL Scheme to Info.plist like this:

• In SecondApp register new URL Scheme in URL Types like this: