单击按钮时如何打开电话设置? [英] How do I open phone settings when a button is clicked?
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问题描述
我正在尝试在应用程序中实现一项功能,该功能在互联网连接不可用时显示警报.警报有两个操作(确定和设置),每当用户点击设置时,我想以编程方式将它们带到手机设置.
I am trying to implement a feature in an App that shows an alert when the internet connection is not available. The alert has two actions (OK and Settings), whenever a user clicks on settings, I want to take them to the phone settings programmatically.
我使用 Swift 和 Xcode.
I am using Swift and Xcode.
推荐答案
Using UIApplication.openSettingsURLString
Swift 5.1 的更新
Update for Swift 5.1
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)
}
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