How to jump to system setting's location service on iOS10? - ios

Before I asked this question I had try:
Use [[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"prefs:root=Privacy&path=LOCATION"]];It's work fine on iOS8 and iOS9,but there is nothing happen on iOS10.
Use [[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];It's work fine on iOS8 and iOS9,too.However,on iOS10,when the app jump to system setting, the system setting exit immediately.
Use [[UIApplication sharedApplication]openURL:url options:#{}completionHandler:nil];It's crashed on iOS8 and iOS9,also,exit immediately on iOS10.
The question is can our app jump to system setting on iOS10? If yes.How?And for [[UIApplication sharedApplication]openURL:url options:#{}completionHandler:nil];what's the optionsmeans?We must code something for the options?

For some time now, apps have only been permitted to open their own settings pane in the settings app. There have been various settings URLs that have worked in the past, but recently Apple has been rejecting apps that use these URLS.
You can open your own application's settings:
if let url = URL(string:UIApplicationOpenSettingsURLString) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
Or in Objective-C
NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if (url != nil) {
[[UIApplication sharedApplication] openURL:url options:[NSDictionary new] completionHandler:nil];
}
If you are targeting version of iOS earlier than 10 then you may prefer to use the older, deprecated, but still functional method:
NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if (url != nil) {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wdeprecated-declarations"
[[UIApplication sharedApplication] openURL:url];
#pragma clang diagnostic pop
}

Note:I use this method for a long time and everyting goes well,but today(2018-9-14),I had been rejected.
Here is my previous answer,do not use this method forever:
CGFloat systemVersion = [[UIDevice currentDevice].systemVersion floatValue];
if (systemVersion < 10) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"App-Prefs:root=Privacy&path=LOCATION"]];
}else{
[[UIApplication sharedApplication]openURL:[NSURL URLWithString:#"App-Prefs:root=Privacy&path=LOCATION"]
options:[NSDictionary dictionary]
completionHandler:nil];
}
Now I use this way:
if (#available(iOS 10.0, *)) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString] options:[NSDictionary dictionary] completionHandler:nil];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
}

Note :- this solution will not be useful for ios10 onwards
Dont forget to add URL schemes :-
Go to Project settings --> Info --> URL Types --> Add New URL Schemes-->URL Schemes = prefs
after that Use this url :-
let settingUrl = URL(string: "App-Prefs:root=Privacy&path=LOCATION")
And open using :-
if #available(iOS 10.0, *) {
UIApplication.shared.open(settingUrl) {
(isOpen:Bool) in
if !isOpen {
debugPrint("Error opening:\(settingUrl.absoluteString)")
// show error
}
}
}else{
if UIApplication.shared.canOpenURL(settingUrl) {
UIApplication.shared.open(settingUrl, completionHandler: { (success) in
print("Settings opened: \(success)") // Prints true
})
}
}
Enjoy :)..this worked for me.

Thanks to this guy. I figure out this URL Scheme Prefs:root=Privacy&path=LOCATION is only available in Today Widget, but no use in containing app.
In Today Widget, you can try this:
[self.extensionContext openURL:[NSURL URLWithString:#"Prefs:root=Privacy&path=LOCATION"] completionHandler:nil];
More about system URL Schemes, you can see here.
This all I got. Hope it will help you.

You can also open your app's setting center by opening"App-Prefs:root=your app bundle id". This will be easy for user to change setting for your app.

This works for me. iOS7 ~ iOS11
But if you are uing the iOS11, you can only jump to the app's setting page 😥
NSURL *url1 = [NSURL URLWithString:#"App-Prefs:root=Privacy&path=LOCATION"];
NSURL *url2 = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if (#available(iOS 11.0, *)) {
if ([[UIApplication sharedApplication] canOpenURL:url2]){
[[UIApplication sharedApplication] openURL:url2 options:#{} completionHandler:nil];
}
} else {
if ([[UIApplication sharedApplication] canOpenURL:url1]){
if (#available(iOS 10.0, *)) {
[[UIApplication sharedApplication] openURL:url1 options:#{} completionHandler:nil];
} else {
[[UIApplication sharedApplication] openURL:url1];
}
}
}

Related

can you force ios to launch safari other than any other browser or even give an option?

Is there a way to launch safari only? I know in order to send an intent and have ios to handle it we can do [[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"http://"]];, but if user has other browser installed (i.e. chrome), there's no guarantee safari will be used.
Reason I want to use safari is that I'm trying to have safari to handle certificate authentication for me, and according to here, only system app has permission to do so
try this
//initially we need to check safari is installed or not in our device
NSURL *url = [NSURL URLWithString:#"safari://"];
UIApplication *application = [UIApplication sharedApplication];
if ([application canOpenURL:url]) {
// if success again need to validate the our calling URL.
NSURL *linkURL = [NSURL URLWithString:#"https://iostree.wordpress.com/2017/07/29/launch-safari-from-ios-app/"];
if ([application canOpenURL:linkURL]) {
if ([application respondsToSelector:#selector(openURL:options:completionHandler:)]) {
[application openURL:linkURL options:#{}
completionHandler:^(BOOL success) {
NSLog(#"success");
}];
}
}
}else{
NSLog(#"safari is not installed");
}

Are we allowed to programmatically open iOS WiFi settings in iOS 10? [duplicate]

I was using prefs:root=WIFI url scheme in my app with prefs entered in info.plist to open directly the iOS settings application in Wi-Fi settings and it was working great on iOS 9 but it does not work anymore on iOS 10.
Does anyone know if this is just a regression in the first developer preview or the way to open Wi-Fi settings has changed in iOS 10 or it is not allowed anymore?
Just so it's explicit: Apple does not allow this. It's possible your app will make it through anyway, but this is the same as using any other undocumented API.
Here is the full list of supported Apple URL schemes.
Here's a thread where Apple confirms that "any Apple URL schemes that are not officially documented should be considered private API."
SWIFT 3.0:- working in iOS 10
#IBAction func openWifiSetting(_ sender: AnyObject) {
let url = URL(string: "App-Prefs:root=WIFI") //for WIFI setting app
UIApplication.shared.openURL(url!)
}
My app is also using that api. Unfortunately apple disable this on iOS 10. Here's my solution: below iOS 10, it can still open Setting App. on iOS 10, it will go to a subpage(Cellular Data access) of Setting App, you can back to setting page by one click. I decide to keep it. because it's still convenient than user manually open Setting App.
NSURL *url = [NSURL URLWithString:#"prefs:root=WIFI"];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
[[UIApplication sharedApplication] openURL:url];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
}
try this for objective c in iOS 10
NSURL *url = [NSURL URLWithString:#"prefs:root=WIFI"];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
[[UIApplication sharedApplication] openURL:url];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"App-Prefs:root=WIFI"]];
}
Using "App-Prefs:root" instead of "prefs:root"
iOS 10, to open your apps settings:
if let settingsURL = URL(string: UIApplicationOpenSettingsURLString) {
UIApplication.shared.openURL(settingsURL)
}
This works fine on iOS 10,
Go to Targets --> (Application) --> Info --> URL Types --> +
In the URL Schemes write
prefs
Then Call,
- (void)openWifiSettings
{
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:#"prefs:root=WIFI"]]) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"prefs:root=WIFI"]];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"App-Prefs:root=WIFI"]];
}
}

Method openURL:options:completionHandler compatibility in objective c

I'm using the method openURL:options:completionHandler:, it turns out that in iOS 10 works fine, but I'm also interested in my app is compatible with the old iOS 9, but xcode gives me a
NSException:
-[UIApplication openURL:options:completionHandler:]:
Unrecognized selector send to instance
There any way make it work in iOS 9 also?
Thank for the possible response!
The new UIApplication method openURL:options:completionHandler:, which
is executed asynchronously and calls the specified completion handler
on the main queue (this method replaces openURL:)
This is under Additional Framework Changes > UIKit at: https://developer.apple.com/library/prerelease/content/releasenotes/General/WhatsNewIniOS/Articles/iOS10.html
you need use it like this:-
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
New method in iOS 10:
- (void)openURL:(NSURL*)url options:(NSDictionary<NSString *, id> *)options
completionHandler:(void (^ __nullable)(BOOL success))completion
Read Doc here:
https://developer.apple.com/library/prerelease/content/releasenotes/General/WhatsNewIniOS/Articles/iOS10.html
The new UIApplication method openURL:options:completionHandler:, which
is executed asynchronously and calls the specified completion handler
on the main queue (this method replaces openURL:).
For below iOS 10:
[[UIApplication sharedApplication] openURL:URL];//URL is NSURL
You can use below code:
UIApplication *application = [UIApplication sharedApplication];
NSURL *URL = [NSURL URLWithString:strUrl];
if([[UIDevice currentDevice].systemVersion floatValue] >= 10.0){
if ([application respondsToSelector:#selector(openURL:options:completionHandler:)]) {
[application openURL:URL options:#{}
completionHandler:^(BOOL success) {
NSLog(#"Open %#: %d",scheme,success);
}];
} else {
BOOL success = [application openURL:URL];
NSLog(#"Open %#: %d",scheme,success);
}
}
else{
bool can = [application canOpenURL:URL];
if(can){
[application openURL:URL];
}
}
Also need to set LSApplicationQueriesSchemes in plist if not set:
Like,
<key>LSApplicationQueriesSchemes</key>
<array>
<string>urlscheme1</string>
<string>urlscheme2</string>
</array>
Also read answer here: https://stackoverflow.com/a/40042291/5575752
You can also use it like as below for obj-c.
[[UIApplication sharedApplication] openURL: url options:#{} completionHandler:nil];

Skype URI's not Working in iOS 9

I had implemented a module for opening Skype app for various modules chat ,call, video call.It was working till iOS 8.
Below is link is followed for integration
https://msdn.microsoft.com/en-us/library/dn745885.aspx
But it stopped working in iOS 9 now.
The below code is just opening the App-store searching Skype even when Skype is installed
BOOL installed = [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:#"skype:"]];
if(installed)
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:#"skype:%#?chat",dataSource[indexPath.section]]]];
}
else
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"https://itunes.apple.com/in/app/skype-for-iphone/id304878510?mt=8"]];
}
Any Alternative for this?Please guide.Thanks
Below i am sharing the image which worked for me.
I added the key LSApplicationQueriesSchemes in Info.plist file for skype
BOOL installed = [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:#"skype:"]];
if(installed)
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:#"skype:%#?chat",dataSource[indexPath.section]]]];
}
else
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"https://itunes.apple.com/in/app/skype-for-iphone/id304878510?mt=8"]];
}

How to call a phone number from ios app? [duplicate]

This question already has answers here:
What’s the correct URL for placing a call on an iPhone?
(5 answers)
Closed 9 years ago.
I'm trying to call a phone number from ios app using:
It's not working, although the method gets called:
-(IBAction)callPhone:(id)sender {
NSString *phoneCallNum = [NSString stringWithFormat:#"tel://%#",listingPhoneNumber ];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneCallNum]];
NSLog(#"phone btn touch %#", phoneCallNum);
}
NSLog output: phone btn touch tel://+39 0668806972
your code is correct. did you check in real device. this function will not work in simulator.
try this also,
NSString *phNo = #"+919876543210";
NSURL *phoneUrl = [NSURL URLWithString:[NSString stringWithFormat:#"telprompt:%#",phNo]];
if ([[UIApplication sharedApplication] canOpenURL:phoneUrl]) {
[[UIApplication sharedApplication] openURL:phoneUrl];
} else
{
calert = [[UIAlertView alloc]initWithTitle:#"Alert" message:#"Call facility is not available!!!" delegate:nil cancelButtonTitle:#"ok" otherButtonTitles:nil, nil];
[calert show];
}
** Swift 3 version**
if let url = URL(string: "telprompt:\(phoneNumber)") {
if UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(call, options: []) { result in
// do something with result
}
}
}
Telephony does not work on simulators/iPod/iPad, you will require to run the app on an iPhone with active sim card.
Also the URL scheme to invoke the telephony application is tel:<phone_number>. Refer Apple docs.
Ideally, you should check if the device is having the telephony module and then perform the openURL: call. Use this code to perform the check,
if([[UIApplication sharedApplication] canOpenURL:callUrl]) {
[[UIApplication sharedApplication] openURL:callUrl];
}
else {
//Show error message to user, etc.
}
Use following method to make call:
NSString *phoneNumber = [#"tel://" stringByAppendingString:number];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
You can try like below.
NSString *phoneNumber = [#"tel://" stringByAppendingString:Number];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
Hope it helps you.
NSString *phoneNumber = [#"tel://" stringByAppendingString:#"9414481799"];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
This will only run on device.

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