答案 0 :(得分:0)
Apple将拒绝您提交的应用程序。因为这是专用API。在下面使用以下功能。参考:How do I open phone settings when a button is clicked?
guard let settingsUrl = URL(string: UIApplication.openSettingsURLString) else {
return
}
if UIApplication.shared.canOpenURL(settingsUrl) {
UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
print("Settings opened: \(success)") // Prints true
})
}