单击按钮时如何打开手机设置? [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.
我是使用夫特和Xcode中。
I am using Swift and Xcode.
推荐答案
使用 UIApplicationOpenSettingsURLString
更新 Swift 3
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)
}
Swift 2.x
override func viewDidAppear(animated: Bool) {
var alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .Alert)
var settingsAction = UIAlertAction(title: "Settings", style: .Default) { (_) -> Void in
let settingsUrl = NSURL(string: UIApplicationOpenSettingsURLString)
if let url = settingsUrl {
UIApplication.sharedApplication().openURL(url)
}
}
var cancelAction = UIAlertAction(title: "Cancel", style: .Default, handler: nil)
alertController.addAction(settingsAction)
alertController.addAction(cancelAction)
presentViewController(alertController, animated: true, completion: nil)
}
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