Allow specific app for opening links
Chisan Gelu
Right2TheV0id
Alan Hendry
Alan Hendry
(2.51 is in Beta testing and includes Func to make it easier to call functions).
Chisan Gelu
How about opening settings for an explicit app.....????I tried this ....
app.SendIntent( "com.myname.distanta", null, "android.intent.action.APPLICATION_PREFERENCES");but didn.t work ....
Chisan Gelu
Activity Action: Show the "Open by Default" page in a particular application's details page.
In some cases, a matching Activity may not exist, so ensure you safeguard against this.
Input: The Intent's data URI specifies the application package name to be shown, with the "package" scheme. That is "package:com.my.app".
Output: Nothing.
Constant Value: "android.settings.APP_OPEN_BY_DEFAULT_SETTINGS"
Chisan Gelu
After the user understands what your app is asking them to do, make the request. To do so, invoke an intent that includes the ACTION_APP_OPEN_BY_DEFAULT_SETTINGS intent action, and a data string matching package:com.example.pkg for the target app, as shown in the following code snippet:
val intent = Intent(Settings.ACTION_APP_OPEN_BY_DEFAULT_SETTINGS,
Uri.parse("package:${context.packageName}"))
context.startActivity(intent)
When the intent is invoked, users see a settings screen called Open by default. This screen contains a radio button called Open supported links, as shown in figure 1.
When the user turns on Open supported links, a set of checkboxes appear under a section called Links to open in this app. From here, users can select the domains that they want to associate with your app. They can also select Add link to add domains, as shown in figure 2. When users later select any link within the domains that they add, the link opens in your app automatically.
