I know there are lot of topics about this problem but is there a solution? I have not found official documentation for that.
My problem is I need to redirect my code on storage settings like this : (work in iOS 9)
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
UIApplication.shared.openURL(url as URL)
}
But since ios10 this method don't work, so Is there an alternative? I saw SnapChat, Google Maps redirect their apps to different part on settings (not the main screen of settings) so I think there is a solution.
I already implement the url scheme in info.plist but it's still not work
I tried this method but same issue too
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
if #available(iOS 10.0, *) {
UIApplication.shared.open(URL.init(string:"Prefs:root=General&path=STORAGE_ICLOUD_USAGE/DEVICE_STORAGE")!, options: [UIApplicationOpenURLOptionUniversalLinksOnly:true], completionHandler:{(success: Bool?) -> Void in})
} else {
UIApplication.shared.openURL(url as URL)
}
}
Thank in advance.
Since iOS 10, it's not possible to open the Settings app from a third party app. The only settings that are allowed to be opened are Keyboard setting but only by a custom keyboard extension and your own application settings.
More details: here
Note: Even for iOS 9, using the URL string that is mentioned in the question can lead to app rejection as it violates iOS App Review Guidelines.
Related
Is there a way to open the Apple Reminders app programmatically?
I have tried in Swift 5+ and iOS 13, you need to use the following url scheme.(Earlier x-apple-reminder:// was working but now it isn't)
x-apple-reminderkit://
if let url = URL(string: "x-apple-reminderkit://"), UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url, options: [:]) { (isDone) in
}
}
It will open the iOS Reminder App from your application.
Hope it may be helpful to other.
Try This
let url = URL(string: "x-apple-reminder://")
UIApplication.shared.open(url!, options: [:]) { (finish) in
}
Reference for all apple apps scheme :
https://ios.gadgethacks.com/news/always-updated-list-ios-app-url-scheme-names-0184033/
and if you use x-apple-reminderkit://today, you can open the Today "view" in Reminders.
Still trying to figure out how to open a specific list or the Flagged view, though.
In order to open instagram app with certain post I'm using following code:
func instaOpen(_ postId: String, _ postUrl: String){
let appURL = URL(string: "instagram://media?id=\(postId)")!
if UIApplication.shared.canOpenURL(appURL) {
UIApplication.shared.open(appURL, options: [:], completionHandler: nil)
} else {
// if Instagram app is not installed, open URL inside Safari
let webURL = URL(string: postUrl)!
let svc = SFSafariViewController(url: webURL)
present(svc, animated: true, completion: nil)
}
}
When instaOpen function called – instagram app opens, but login prompt forcefully pops over. Not matter what you do - close it or proceed with login, the queried post simply won't open(see gif).
This started happening recently, after I've updated my app and pushed deployment target to iOS12.
I do have instagram listed in my LSApplicationQueriesSchemes as well as I'm 100% positive that correct mediaID is being passed to instaOpen func (the code worked previously).
Let me know if there's any suggestions on how to fix this and actually open instagram post in instagram app.
Updated - Facebook developer fixed the issue.
Its instagram bug you can follow its progress from https://developers.facebook.com/support/bugs/290173615155052/?disable_redirect=0
Probably a bug, as that feature works on Android.
i manage to "fix" the problem on a pwa app using the Instagram web app.
let appURL = URL(string: "https://www.instagram.com/p/\(postId)")!
//https://www.instagram.com/p/insert here media id
I don't know how to check if app is installed or not on phone! Or when App is installed, open the app, otherwise open the Appstore link to download the app. I'm using swift 3. I want to do it using app name or bundle identifier.
Thank You!
func openApp(appName:String) {
let appName = "instagram"
let appScheme = "\(appName)://app"
let appUrl = URL(string: appScheme)
if UIApplication.shared.canOpenURL(appUrl! as URL) {
UIApplication.shared.open(appUrl!)
} else {
print("App not installed")
}
}
After looking into so many answers, i am writing a common code which will help for new users. If you have two mobile apps as App1 and App2, if you want to check in App2 that App1 is already installed in his device or not, here is code below.
In App1 add this property in info.plist file
<key>CFBundleURLTypes</key>
<array>
<dict>
<key>CFBundleURLName</key>
<string>com.companyName.App1</string>
<key>CFBundleURLSchemes</key>
<array>
<string>App1</string>
</array>
</dict>
</array>
In App2 add this property in info.plist file
<key>LSApplicationQueriesSchemes</key>
<array>
<string>App1</string>
</array>
In App2 write the method to check if app is installed or not on phone! Or when App is installed, open the app, otherwise open the Appstore link to download the app as below.
func checkAndOpenApp(){
let app = UIApplication.shared
let appScheme = "App1://app"
if app.canOpenURL(URL(string: appScheme)!) {
print("App is install and can be opened")
let url = URL(string:appScheme)!
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
} else {
print("App in not installed. Go to AppStore")
if let url = URL(string: "https://apps.apple.com/us/app/App1/id1445847940?ls=1"),
UIApplication.shared.canOpenURL(url)
{
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
}
}
}
I hope it will help some one.
Between the other answers and their comments, I'm getting the impression that the asker wants to be able to see if any given app is installed.
Beginning with iOS 9.0, that is not possible.
Apps for iOS 9 and later must have a list of requested URL schemes in the Info.plist before being allowed to use canOpenURL:. This is to protect user privacy, as advertisers were abusing this functionality in an arguably invasive fashion. (See this excellent post for more details on those changes.)
Of course, that list is static and cannot be changed after build time or submission to the App Store. If Apple doesn't like the ones you chose, they have every right to reject it.
I'm afraid that what you're asking isn't possible within reason for iOS 9.0 and later.
Edit: I also want to make clear that an app's URL scheme may not necessarily match nicely with its name. (This is more of an issue of a badly named constant than a functional issue.) There used to be a giant list of known URI schemes with documentation for each, but poignantly and fittingly enough, it seems that the wiki hosting it has shut down.
Swift 4.1. One developer can have more than one app on AppStore. So, I need to check if user has installed other apps or not by the same developer. I had Bundle ID's of other apps. Although you can use Appname instead of Bundle Id. So I followed the following steps.
In your current app add LSApplicationQueriesSchemes key of type Array in your info.plist file. Make entry of bundle id or Appname of app there which you want to open from your app.
Other app should have their bundle id or Appname entry in that app URL Scheme.
In your current app check if that app in installed in iPhone or not and can open accordingly.
let app = UIApplication.shared
let bundleID = "some.Bundle.Id://"
if app.canOpenURL(URL(string: bundleID)!) {
print("App is install and can be opened")
let url = URL(string:bundleID)!
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
} else {
print("App in not installed. Go to AppStore")
}
You can also test it from Safari browser. Just type the following in search bar
URL_scheme:// or Bundle_Id://
If app is installed the it will show alert with Appname to open it in app.
This worked for me (Swift 3.0)
Below two inputs should be provided:
<APP URL SCEHME>: The URL Scheme of the app which you want to open
<YOUR APP URL>: The App Itunes URL
func openApp() {
let appURL = NSURL(string: "<APP URL SCHEME>")
if UIApplication.shared.canOpenURL(appURL as! URL) {
print("Opening App...")
}else {
UIApplication.shared.openURL(NSURL(string: "<YOUR APP URL>")! as URL)
}
}
first go to info.plist, add LSApplicationQueriesSchemes add an item and place instagram on that item. Now this code will run perfectly.
let appName = "instagram"
let appScheme = "\(appName)://"
let appUrl = URL(string: appScheme)
if UIApplication.shared.canOpenURL(appUrl! as URL)
{
UIApplication.shared.open(appUrl!)
} else {
print("App not installed")
}
I need to open Zattoo app from my app (on a button click event)
what I have done is
let url:NSURL? = NSURL(string: "zattoo://")
if UIApplication.sharedApplication().canOpenURL(url!) {
UIApplication.sharedApplication().openURL(url!)
} else {
print("App not installed")
//redirect to safari because the user doesn't have Zattoo App installed
UIApplication.sharedApplication().openURL(NSURL(string: "https://itunes.apple.com/de/app/zattoo-tv-app-sports-news/id423779936?l=en")!)
}
canOpenURL() always returns me the false (even Zattoo app is installed on my device) hence the code in else is executed always.
but if removing the check and just executing the
let url:NSURL? = NSURL(string: "zattoo://")
UIApplication.sharedApplication().openURL(url!)
It's opening the the Zattoo app perfectly. Strange!
What I am doing wrong?
In iOS9 you should register custom schemes you want to use to open other apps. It should be stored as array of strings (custom schemes) with LSApplicationQueriesSchemes key in Info.plist file. From official documentation ( https://developer.apple.com/library/ios/documentation/General/Reference/InfoPlistKeyReference/Articles/LaunchServicesKeys.html#//apple_ref/doc/uid/TP40009250-SW14 ) :
LSApplicationQueriesSchemes (Array - iOS) Specifies the URL schemes
you want the app to be able to use with the canOpenURL: method of the
UIApplication class. For each URL scheme you want your app to use with
the canOpenURL: method, add it as a string in this array. Read the
canOpenURL: method description for important information about
declaring supported schemes and using that method
There is an example how it can be used: https://github.com/gatzsche/LSApplicationQueriesSchemes-Working-Example
Try this :-
if let url = NSURL(string: "zattoo://") {
let canOpen = UIApplication.sharedApplication().canOpenURL(url)
}
Hope it helps.
So I am about to launch an app to the App Store. My issue is that I have a rate my app please button but I do not know the right code to insert there.
My fiends tried this on their app and said it was no good:
Does anybody know how I can fix this issue?
let iTunesReviewPageLink = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?id=1073785561&pageNumber=0&sortOrdering=2&type=Purple+Software&mt=8"
// Go directly to review page on App Store
if let url = NSURL(string: iTunesReviewPageLink) {
UIApplication.sharedApplication().openURL(url)
}
The only unknown thing is the ID, right? You can see the ID of your app before it is published - once you set it up in iTunes Connect.
If your app is not yet released, you haven't got an App Store link. So that's impossible.
In order to implement this functionality when your app is released, you can use the following code:
Swift 2
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = NSURL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.sharedApplication().canOpenURL(reviewsURL!) {
UIApplication.sharedApplication().openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
Swift 3
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = URL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.shared.canOpenURL(reviewsURL!) {
UIApplication.shared.openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
EDIT:
This links works in iOS 8 and 9 and links directly to reviews page of the app in App Store application.
I am not sure about iOS 7. Probably for iOS 7 you need to use different link.