我试图在一个应用程序中实现一个功能,当互联网连接不可用时显示警报。 警报有两个动作(确定和设置),每当用户单击设置,我想以编程方式将他们带到电话设置。

我使用Swift和Xcode。


当前回答

UIApplication.open(_:options:completionHandler:)只能在主线程中使用

解决方案:

if let appSettings = URL(string: UIApplication.openSettingsURLString + Bundle.main.bundleIdentifier!) {
  if UIApplication.shared.canOpenURL(appSettings) {
    DispatchQueue.main.async {
        UIApplication.shared.open(appSettings)
    }
  }
}

其他回答

我看过这行代码

UIApplication.sharedApplication() .openURL(NSURL(string:"prefs:root=General")!)

是不工作,它没有为我在ios10/ Xcode 8,只是一个小的代码差异,请替换这个

UIApplication.sharedApplication().openURL(NSURL(string:"App-Prefs:root=General")!)

Swift3

UIApplication.shared.openURL(URL(string:"prefs:root=General")!)

替换为

UIApplication.shared.openURL(URL(string:"App-Prefs:root=General")!)

希望能有所帮助。 欢呼。

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)
}

斯威夫特4

如果这是你想要的,这可能会占用你应用程序的特定设置。

UIApplication.shared.openURL(URL(string: UIApplicationOpenSettingsURLString)!)

在iOS 10.3上,App-Specific URL Schemes的第一个响应对我来说很有效。

if let appSettings = URL(string: UIApplicationOpenSettingsURLString + Bundle.main.bundleIdentifier!) {
    if UIApplication.shared.canOpenURL(appSettings) {
      UIApplication.shared.open(appSettings)
    }
  }

在ios10/ Xcode 8模拟器:

UIApplication.shared.openURL(URL(string:UIApplicationOpenSettingsURLString)!)

作品

UIApplication.shared.openURL(URL(string:"prefs:root=General")!)

没有。