单击按钮时如何打开电话设置？

一尘不染

单击按钮时如何打开电话设置？

swift

我正在尝试在应用程序中实现一项功能，当互联网连接不可用时会显示警报。该警报具有两个操作（“确定”和“设置”），每当用户单击设置时，我都希望以编程方式将其设置为电话设置。

我正在使用Swift和Xcode。

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2020-07-07

共1个答案

一尘不染

使用 UIApplication.openSettingsURLString

Swift 5.1更新

 override func viewDidAppear(_ animated: Bool) {
    let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)

    let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in

        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
            })
        }
    }
    alertController.addAction(settingsAction)
    let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
    alertController.addAction(cancelAction)

    present(alertController, animated: true, completion: nil)
}

斯威夫特4.2

override func viewDidAppear(_ animated: Bool) {
    let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)

    let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in

        guard let settingsUrl = URL(string: UIApplicationOpenSettingsURLString) else {
            return
        }

        if UIApplication.shared.canOpenURL(settingsUrl) {
            UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
                print("Settings opened: \(success)") // Prints true
            })
        }
    }
    alertController.addAction(settingsAction)
    let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
    alertController.addAction(cancelAction)

    present(alertController, animated: true, completion: nil)
}

2020-07-07