问题
I want to open location service screen programmatically to turn on service.
回答1:
Location services App-Prefs:root=Privacy&path=LOCATION
worked for me. When I tested on a device and not a simulator.
I won't list the things I tried that did not work, it's a long list.
Usage example that assumes either location services are disabled or permission is denied or not determined:
if !CLLocationManager.locationServicesEnabled() {
if let url = URL(string: "App-Prefs:root=Privacy&path=LOCATION") {
// If general location settings are disabled then open general location settings
UIApplication.shared.openURL(url)
}
} else {
if let url = URL(string: UIApplicationOpenSettingsURLString) {
// If general location settings are enabled then open location settings for the app
UIApplication.shared.openURL(url)
}
}
回答2:
I have tried all the above answers,it's not working on iOS11..it just opens settings page and not the app settings .. Finally this works..
UIApplication.shared.open(URL(string:UIApplicationOpenSettingsURLString)!)
Swift 4.2:
UIApplication.shared.open(URL(string:UIApplication.openSettingsURLString)!)
Refer: https://developer.apple.com/documentation/uikit/uiapplicationopensettingsurlstring?language=swift
回答3:
You can open it directly like using below code,
But first set URL Schemes
in Info.plist's URL Type Like:
Then write below line at specific event:
In Objective - C :
[[UIApplication sharedApplication] openURL:
[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
In Swift :
UIApplication.sharedApplication().openURL(NSURL(string: "prefs:root=LOCATION_SERVICES")!)
Hope this will help you.
回答4:
SWIFT 4 tested:
Only way to avoid getting rejected and open Location Preferences of own app is:
if let bundleId = Bundle.main.bundleIdentifier,
let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
回答5:
Swift 4.2
Go straight to YOUR app's settings like this. Don't forget to put in your bundle identifier -
if let bundleId = Bundle.main.bundleIdentifier,
let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)")
{
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
回答6:
Step 1: Click on project name >> target>> info >> url Types
Step 2:
-(IBAction)openSettingViewToEnableLocationService:(id)sender
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
}
回答7:
First:
Add URL
Go to Project settings --> Info --> URL Types --> Add New URL Schemes
See image below:
Second:
Use below code to open Location settings:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
referred from: https://stackoverflow.com/a/35987082/5575752
回答8:
🚨🚨
Do you want to be safe? use UIApplicationOpenSettingsURLString
, which will open the app settings, without deep-link.
Using App-prefs
your app will be rejected, as many sub comments said.
https://github.com/mauron85/cordova-plugin-background-geolocation/issues/394
回答9:
If you set locationManager.startUpdatingLocation() and you have disabled on your iphone, it automatically show you an alertView with the option to open and activated location.
回答10:
Actually there's much simpler solution to that. It'll show your app settings with loction services/camera access, etc.:
func showUserSettings() {
guard let urlGeneral = URL(string: UIApplicationOpenSettingsURLString) else {
return
}
UIApplication.shared.open(urlGeneral)
}
回答11:
After adding prefs as a url type, use the following code to go directly to the location settings of an application.
if let url = URL(string: "App-prefs:root=LOCATION_SERVICES") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
来源:https://stackoverflow.com/questions/37654132/how-to-open-location-services-screen-from-setting-screen