Opening the Settings app from another app

问题

Okay, I know that there are many question about it, but they are all from many time ago.

So. I know that it is possible because the Map app does it.

In the Map app if I turn off the localization for this app, it send me a message, and if I press okay, the \"Settings App\" will be open. And my question is, how is this possible? How can I open the \"Setting app\" from my own app?

Basically I need to do the same thing, if the user turn off the location for my app, then I\'ll show him a message saying something that will open the \"Setting app\"

回答1:

As mentioned by Karan Dua this is now possible in iOS8 using UIApplicationOpenSettingsURLString see Apple's Documentation.

Example:

Swift 4.2

UIApplication.shared.open(URL(string: UIApplication.openSettingsURLString)!)

In Swift 3:

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

In Swift 2:

UIApplication.sharedApplication().openURL(NSURL(string:UIApplicationOpenSettingsURLString)!)

In Objective-C

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];

Prior to iOS 8:

You can not. As you said this has been covered many times and that pop up asking you to turn on location services is supplied by Apple and not by the App itself. That is why it is able to the open the settings application.

Here are a few related questions & articles:

is it possible to open Settings App using openURL?

Programmatically opening the settings app (iPhone)

How can I open the Settings app when the user presses a button?

iPhone: Opening Application Preferences Panel From App

Open UIPickerView by clicking on an entry in the app's preferences - How to?

Open the Settings app?

iOS: You’re Doing Settings Wrong

回答2:

From @Yatheeshaless's answer:

You can open settings app programmatically in iOS8, but not in earlier versions of iOS.

Swift:

   UIApplication.sharedApplication().openURL(NSURL(string:UIApplicationOpenSettingsURLString)!)

Swift 4:

if let url = NSURL(string: UIApplicationOpenSettingsURLString) as URL? {
    UIApplication.shared.openURL(url)
}

Swift 4.2 (BETA):

if let url = NSURL(string: UIApplication.openSettingsURLString) as URL? {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Objective-C:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];

回答3:

YES!! you can launch Device Settings screen, I have tested on iOS 9.2

Step 1. we need to add URL schemes

Go to Project settings --> Info --> URL Types --> Add New URL Schemes

Step 2. Launch Settings programmatically Thanks to @davidcann

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs://"]];

Also we can launch sub-screens like Music, Location etc. as well by just using proper name

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=MUSIC"]];

See this full name list here shared by Henri Normak

Update:

As per the comment everyone wants to know what happens after this change to my application submission status?

So YES!! I got successful update submission and application is available on store without any complain.

Just to confirm, I Just downloaded this morning and disabled Location services, and then started the app, which asked me for location permission and then my alert popup was there to send me on settings -> location services page --> Enabled --> That's it!!

![NOTICE: Your app might be rejected ... even if it's approved it can be rejected in future version if you use this method...]4

回答4:

~~You can use this on iOS 5.0 and later:~~ This no longer works.

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs://"]];

回答5:

iOS 10 update

Apple changed the method to open async on the main thread. However, from now it is only possible to open the app settings in native settings.

[[UIApplication sharedApplication] openURL:url options:@{} completionHandler:nil];

iOS 9 update

It is now possible to go directly to sub-settings menu. However, a URL scheme has to be created. It can be done using two ways:

XCode - You will find it in Target, Info, URL Scheme. Then, just type prefs.
Directly adding to *-Info.plist. Add the following: <key>CFBundleURLTypes</key> <array> <dict> <key>CFBundleTypeRole</key> <string>Editor</string> <key>CFBundleURLSchemes</key> <array> <string>prefs</string> </array> </dict> </array>

Then the code:

Swift

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

Objective-c

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=General&path=Keyboard"]];

回答6:

In Swift 3 / iOS 10+ this now looks like

if let url = URL(string: "App-Prefs:root=LOCATION_SERVICES") {
    UIApplication.shared.open(url, completionHandler: .none)
}

回答7:

Swift 3:

guard let url = URL(string: UIApplicationOpenSettingsURLString) else {return}
if #available(iOS 10.0, *) {
  UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
  // Fallback on earlier versions
  UIApplication.shared.openURL(url)
}

回答8:

In Swift 3 all I needed is this (here for example redirect to my app notifications):

if let url = URL(string: "App-Prefs:root=NOTIFICATIONS_ID&path=your app bundleID") {
    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, completionHandler: .none)
    } else {
        // Fallback on earlier versions
    }
}

Source: phynet gist.

This worked with me only when settings is in background. It will redirect you to your app notification settings but if settings wasn't running in the background it will just redirect you to notification settings in general.

回答9:

UIApplicationOpenSettingsURLString this will only work if you have previously allowed for any permission. For example Location, Photo, Contact, Push notification access. So if you have not such permission(s) from the user:

If iOS 10 or above,

It will open the Settings but then crash it. The reason, there's nothing in settings for your app.

Below code will open your application settings inside the iOS Setting.

NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
    [[UIApplication sharedApplication] openURL:url];
}

Due to device unavailability, I couldn't check this on iOS < 10.

Also, I could find below code from some gist and it works fine on iOS 10 as well. But I am not sure if this will approve by Apple review team or not.

https://gist.github.com/johnny77221/bcaa5384a242b64bfd0b8a715f48e69f

回答10:

You can use the below code for it.

[[UIApplication sharedApplication]openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];

回答11:

SWIFT 4.0

'openURL' was deprecated in iOS 10.0: Please use openURL:options:completionHandler: instead

UIApplication.shared.open(URL.init(string: UIApplicationOpenSettingsURLString)! , options: [:], completionHandler: nil)

回答12:

Swift 4

I prefer to open setting in a safer way,

if let settingUrl = URL(string:UIApplicationOpenSettingsURLString) {

    UIApplication.shared.open(settingUrl)
}
else {
    print("Setting URL invalid")
}

回答13:

Swift You can use following function to open Settings App with Bluetooth Page

func openSettingsApp(){
    if let settings = NSURL(string: "prefs:root=Bluetooth") {
        UIApplication.sharedApplication().openURL(settings)
    }
}

Again this would not open the App's Settings. This would open settings app with Bluetooth as this is deep linking to bluetooth.

回答14:

Tested with iOS 10. Working

NSArray* urlStrings = @[@"prefs:root=WIFI", @"App-Prefs:root=WIFI"];
for(NSString* urlString in urlStrings){
NSURL* url = [NSURL URLWithString:urlString];
if([[UIApplication sharedApplication] canOpenURL:url]){
    [[UIApplication sharedApplication] openURL:url];
    break;
}
}

Happy Coding :)

回答15:

To add to accepted answer: (from apple developer documentation) "When you open the URL built from this string (openSettingsURLString), the system launches the Settings app and displays the app’s custom settings, if it has any." So, if you want to open settings for your app, create your own Settings.bundle.

回答16:

Add this to your class,

 public class func showSettingsAlert(title:String,message:String,onVC viewController:UIViewController,onCancel:(()->())?){
            YourClass.show2ButtonsAlert(onVC: viewController, title: title, message: message, button1Title: "Settings", button2Title: "Cancel", onButton1Click: {
                if let settingsURL = NSURL(string: UIApplicationOpenSettingsURLString){
                    UIApplication.sharedApplication().openURL(settingsURL)
                }
                }, onButton2Click: {
                    onCancel?()
            })
        }

 public class func show2ButtonsAlert(onVC viewController:UIViewController,title:String,message:String,button1Title:String,button2Title:String,onButton1Click:(()->())?,onButton2Click:(()->())?){
            dispatch_async(dispatch_get_main_queue()) {
                let alert : UIAlertController = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.Alert)

                alert.addAction(UIAlertAction(title: button1Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton1Click?()
                }))

                alert.addAction(UIAlertAction(title: button2Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton2Click?()
                }))

                viewController.presentViewController(alert, animated: true, completion: nil)
            }
        }

Call like this,

YourClass.showSettingsAlert("App would like to access camera", message: "App would like to access camera desc", onVC: fromViewController, onCancel: {
  print("canceled")
})

来源：https://stackoverflow.com/questions/5655674/opening-the-settings-app-from-another-app

标签

ios

objective-c

iphone

application-settings

appsettings