ios 如何从设置屏幕打开定位服务屏幕？
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How to open Location services screen from setting screen?
提问by Monika Patel
I want to open location service screen programmatically to turn on service.
我想以编程方式打开定位服务屏幕以打开服务。
采纳答案by Monika Patel
Step 1: Click on project name >> target>> info >> url Types
第一步：点击项目名称>>目标>>信息>>网址类型
Step 2:
第2步：
-(IBAction)openSettingViewToEnableLocationService:(id)sender
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
}
回答by Vijay Karthik
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..
我已经尝试了上述所有答案，它在 iOS11 上不起作用..它只是打开设置页面而不是应用程序设置..最后这有效..
UIApplication.shared.open(URL(string:UIApplicationOpenSettingsURLString)!)
Swift 4.2:
斯威夫特 4.2：
UIApplication.shared.open(URL(string:UIApplication.openSettingsURLString)!)
Refer: https://developer.apple.com/documentation/uikit/uiapplicationopensettingsurlstring?language=swift
参考：https: //developer.apple.com/documentation/uikit/uiapplicationopensettingsurlstring?language=swift
回答by 3vangelos
SWIFT 4 tested:
SWIFT 4 测试：
Only way to avoid getting rejected and open Location Preferences of own app is:
避免被拒绝并打开自己的应用程序的位置首选项的唯一方法是：
if let bundleId = Bundle.main.bundleIdentifier,
let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
回答by Jigar Tarsariya
You can open it directly like using below code,
您可以像使用下面的代码一样直接打开它，
But first set URL Schemesin Info.plist's URL Type Like:
但首先URL Schemes在 Info.plist 的 URL Type Like 中设置：
Then write below line at specific event:
然后在特定事件下面写下一行：
In Objective - C:
在目标-C 中：
[[UIApplication sharedApplication] openURL:
[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
In Swift:
在斯威夫特：
UIApplication.sharedApplication().openURL(NSURL(string: "prefs:root=LOCATION_SERVICES")!)
Hope this will help you.
希望这会帮助你。
回答by Trev14
Swift 4.2
斯威夫特 4.2
Go straight to YOUR app's settings like this. Don't forget to put in your bundle identifier -
像这样直接进入您的应用程序设置。不要忘记输入您的包标识符 -
if let bundleId = Bundle.main.bundleIdentifier,
let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)")
{
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
回答by Ronak Chaniyara
First:
第一的：
Add URL
添加网址
Go to Project settings --> Info --> URL Types --> Add New URL Schemes
转到项目设置 --> 信息 --> URL 类型 --> 添加新的 URL 方案
See image below:
见下图：
Second:
第二：
Use below code to open Location settings:
使用以下代码打开位置设置：
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
referred from: https://stackoverflow.com/a/35987082/5575752
回答by Estev?o Lucas
Do you want to be safe? use UIApplicationOpenSettingsURLString, which will open the app settings, without deep-link.
你想安全吗？使用UIApplicationOpenSettingsURLString，这将打开应用程序设置，没有深层链接。
Using App-prefsyour app will be rejected, as many sub comments said.
https://github.com/mauron85/cordova-plugin-background-geolocation/issues/394
App-prefs正如许多子评论所说，使用您的应用程序将被拒绝。
https://github.com/mauron85/cordova-plugin-background-geolocation/issues/394
回答by Nicolas Madrid
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.
如果您设置了 locationManager.startUpdatingLocation() 并且您在 iPhone 上禁用了它，它会自动向您显示一个带有打开和激活位置选项的 alertView。
回答by Shakeel Ahmed
Swift 5+
Easy Way Direct Your said application page will open
Swift 5+
Easy Way Direct 您所说的应用程序页面将打开
if let BUNDLE_IDENTIFIER = Bundle.main.bundleIdentifier,
let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(BUNDLE_IDENTIFIER)") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
回答by Mateusz Grzegorzek
Actually there's much simpler solution to that. It'll show your app settings with loction services/camera access, etc.:
实际上有更简单的解决方案。它会显示您的应用设置以及位置服务/相机访问等：
func showUserSettings() {
guard let urlGeneral = URL(string: UIApplicationOpenSettingsURLString) else {
return
}
UIApplication.shared.open(urlGeneral)
}
