Launch XCode UI test with URL launch option - ios

I'd like to write a test for my iOS app that verifies the correct screen content is shown when the app is launched from a custom registered URL scheme.
For example, a user receives an email with a link to myapp://action1/1234. When they tap on this link, my app is launched and the screen displays "1234".
In didFinishLaunchingWithOptions my AppDelegate checks to see if launchOptions?[UIApplicationLaunchOptionsURLKey] exists and takes appropriate action.
How do I write a UI test so that the app launchOptions dictionary contains the expected URL when the app is launched?

You can make this work by leveraging launchEnvironment on XCUIApplication in combination with a few lines of custom code in your target app. Basically you would set a custom launch environment variable for this test, and in your app you would check for that variable and handle it the same way you would handle the existence of the particular UIApplicationLaunchOptionsURLKey you're expecting.
In your app you can check that value via NSProcessInfo.processInfo.environment which returns a dictionary of the environment variables.
(Take note that the default XCTestCase template in Xcode 7 includes a call to XCUIApplication -launch in setup which will terminate any previously running instance, and launch a new instance by default.)
Edit: wanted to add, I've written a little bit more about this in an article on Xcode 7 UI Testing Tips, in case it's helpful.

Related

Open your own application via URL schemes [duplicate]

I tested:
UIControl().sendAction(#selector(URLSessionTask.suspend), to: UIApplication.shared, for: nil)
which is for putting app on background and it works.
How do I put app back on foreground?
I tried:
UIControl().sendAction(#selector(URLSessionTask.resume), to: UIApplication.shared, for: nil)
But eventually it crashes...
Thank you
Update:
Since you've indicated that you're looking for any technical solution, even those not compatible with the App Store or Apple's terms, this should be possible using the Private API LSApplicationWorkspace: openApplicationWithBundleID. Try something like this:
Create a .h file and set up an interface to the LSApplicationWorkspace class and list the required method. You will need to #import "PrivateHeaders.h" in your bridging header.
//
// PrivateHeaders.h
//
#ifndef PrivateHeaders_h
#define PrivateHeaders_h
#interface LSApplicationWorkspace : NSObject
- (bool)openApplicationWithBundleID:(id)arg1;
#end
#endif /* PrivateHeaders_h */
You should then be able to call this function and pass in the Bundle Identifier of your app as an string.
//
// SomeClass.swift
//
import MobileCoreServices
let workspace = LSApplicationWorkspace()
/**
Launch an App given its bundle identifier
- parameter bundleIdentifier: The bundle identifier of the app to launch
- returns: True if app is launched, otherwise false
*/
func openApp(withBundleIdentifier bundleIdentifier: String) -> Bool {
// Call the Private API LSApplicationWorkspace method
return workspace.openApplication(withBundleID: bundleIdentifier)
}
Original:
What you are doing is likely a violation of the iOS Human Interface Guidelines (although the "Don’t Quit Programmatically" is no longer specifically defined), so as the comments have said, it is not suited to the App Store. Regardless, once your app is suspended in this way, I don't expect that there is a way to resume it programmatically, unless you can hook into a Background Operation to run URLSessionTask.resume, but I have not tested it and am unsure whether it can work.
Apps can be launched (and hence brought into the foreground) programmatically from another app or today extension by using a Custom URL Scheme, or via a Push Notification. It isn't possible to launch the app from the Background Operation via a URL Scheme, since it is part of the UIKit framework, which must be run in the main thread.
In summary, I think your best option is to try to use a Notification. This just means that the user will need to click on the notification to bring your app back into the foreground.
Closing/opening the app should be done explicitly by the user. Any other way of closing or opening the app is not supported by Apple and will be rejected when uploaded to app store. iOS Human Interface Guideline states:
Don’t Quit Programmatically
Never quit an iOS application
programmatically because people tend to interpret this as a crash.
However, if external circumstances prevent your application from
functioning as intended, you need to tell your users about the
situation and explain what they can do about it. Depending on how
severe the application malfunction is, you have two choices.
*Display
an attractive screen that describes the problem and suggests a
correction. A screen provides feedback that reassures usersthat
there’s nothing wrong with your application. It puts usersin control,
letting them decide whether they want to take corrective action and
continue using your application or press the Home button and open a
different application
*If only some of your application's features are
not working, display either a screen or an alert when people activate
the feature. Display the alert only when people try to accessthe
feature that isn’t functioning
Just as a follow up to Jordan's excellent answer I want to give an explanation for why your code works in the first place and why that alone will get your app rejected, even without any functionality to make it active again and bring it to the foreground.
As maddy pointed out in a comment, you're basically calling a method from UIApplication's private API. This works due to the Objective-C runtime's dynamic linking. You might wonder "But I am using Swift, what does that have to do with Objective-C?" The answer lies in #selector mechanism. A Selector is basically just a symbol that the Objective-C runtime looks up in a table to get a method it invokes (for you). This is why it's technically not correct to say you "call a method" when you do something like myObjectInstance.someMethod(). The correct way to phrase that would be to "send a message" to the object, because that's what is happening in the runtime. The target-action mechanism is build around that. The sendAction(_: Selector?, to: Any?) method does the same thing. So in effect your code does the following:
Get the symbol that corresponds to URLSessionTask's suspend() method.
Tell the shared instance of UIApplication to invoke the method that it has for that symbol.
Now usually that would result in a crash with the typical "unknown selector sent to instance..." error message. But here, by sure coincidence UIApplication also has a method for that instance (or rather, the runtime also has one of its methods listed in its table for that symbol). You kind of "found" a method that is not declared in its public header. You successfully circumvented a compile-time check for this and invoke a method that is part of a private API. This is explicitly forbidden in the Apple Developer Program License Agreement
Besides all that, I would strongly advise against trying to design an app that way in the first place. As maddy pointed out it's also likely considered to violate the HIGs. Even if you're not trying to do anything malicious and properly explain the feature in your app's description, that won't make Apple let it slide (I assume). Personally, as a user, I'd also find it annoying if the app did something the system already has a specific mechanic for in a different manner, at least in terms of app's coming to background and foreground.
I don't think it can be done without user interaction
The option is you can generate a push notification to tell the user to bring the application to foreground
When the operating system delivers push notification and the target application is not running in the foreground, it presents the notification.
If there is a notification alert and the user taps or clicks the action button (or moves the action slider), the application launches and calls a method to pass in the local-notification object or remote-notification payload.

Put app on foreground programmatically on Swift

I tested:
UIControl().sendAction(#selector(URLSessionTask.suspend), to: UIApplication.shared, for: nil)
which is for putting app on background and it works.
How do I put app back on foreground?
I tried:
UIControl().sendAction(#selector(URLSessionTask.resume), to: UIApplication.shared, for: nil)
But eventually it crashes...
Thank you
Update:
Since you've indicated that you're looking for any technical solution, even those not compatible with the App Store or Apple's terms, this should be possible using the Private API LSApplicationWorkspace: openApplicationWithBundleID. Try something like this:
Create a .h file and set up an interface to the LSApplicationWorkspace class and list the required method. You will need to #import "PrivateHeaders.h" in your bridging header.
//
// PrivateHeaders.h
//
#ifndef PrivateHeaders_h
#define PrivateHeaders_h
#interface LSApplicationWorkspace : NSObject
- (bool)openApplicationWithBundleID:(id)arg1;
#end
#endif /* PrivateHeaders_h */
You should then be able to call this function and pass in the Bundle Identifier of your app as an string.
//
// SomeClass.swift
//
import MobileCoreServices
let workspace = LSApplicationWorkspace()
/**
Launch an App given its bundle identifier
- parameter bundleIdentifier: The bundle identifier of the app to launch
- returns: True if app is launched, otherwise false
*/
func openApp(withBundleIdentifier bundleIdentifier: String) -> Bool {
// Call the Private API LSApplicationWorkspace method
return workspace.openApplication(withBundleID: bundleIdentifier)
}
Original:
What you are doing is likely a violation of the iOS Human Interface Guidelines (although the "Don’t Quit Programmatically" is no longer specifically defined), so as the comments have said, it is not suited to the App Store. Regardless, once your app is suspended in this way, I don't expect that there is a way to resume it programmatically, unless you can hook into a Background Operation to run URLSessionTask.resume, but I have not tested it and am unsure whether it can work.
Apps can be launched (and hence brought into the foreground) programmatically from another app or today extension by using a Custom URL Scheme, or via a Push Notification. It isn't possible to launch the app from the Background Operation via a URL Scheme, since it is part of the UIKit framework, which must be run in the main thread.
In summary, I think your best option is to try to use a Notification. This just means that the user will need to click on the notification to bring your app back into the foreground.
Closing/opening the app should be done explicitly by the user. Any other way of closing or opening the app is not supported by Apple and will be rejected when uploaded to app store. iOS Human Interface Guideline states:
Don’t Quit Programmatically
Never quit an iOS application
programmatically because people tend to interpret this as a crash.
However, if external circumstances prevent your application from
functioning as intended, you need to tell your users about the
situation and explain what they can do about it. Depending on how
severe the application malfunction is, you have two choices.
*Display
an attractive screen that describes the problem and suggests a
correction. A screen provides feedback that reassures usersthat
there’s nothing wrong with your application. It puts usersin control,
letting them decide whether they want to take corrective action and
continue using your application or press the Home button and open a
different application
*If only some of your application's features are
not working, display either a screen or an alert when people activate
the feature. Display the alert only when people try to accessthe
feature that isn’t functioning
Just as a follow up to Jordan's excellent answer I want to give an explanation for why your code works in the first place and why that alone will get your app rejected, even without any functionality to make it active again and bring it to the foreground.
As maddy pointed out in a comment, you're basically calling a method from UIApplication's private API. This works due to the Objective-C runtime's dynamic linking. You might wonder "But I am using Swift, what does that have to do with Objective-C?" The answer lies in #selector mechanism. A Selector is basically just a symbol that the Objective-C runtime looks up in a table to get a method it invokes (for you). This is why it's technically not correct to say you "call a method" when you do something like myObjectInstance.someMethod(). The correct way to phrase that would be to "send a message" to the object, because that's what is happening in the runtime. The target-action mechanism is build around that. The sendAction(_: Selector?, to: Any?) method does the same thing. So in effect your code does the following:
Get the symbol that corresponds to URLSessionTask's suspend() method.
Tell the shared instance of UIApplication to invoke the method that it has for that symbol.
Now usually that would result in a crash with the typical "unknown selector sent to instance..." error message. But here, by sure coincidence UIApplication also has a method for that instance (or rather, the runtime also has one of its methods listed in its table for that symbol). You kind of "found" a method that is not declared in its public header. You successfully circumvented a compile-time check for this and invoke a method that is part of a private API. This is explicitly forbidden in the Apple Developer Program License Agreement
Besides all that, I would strongly advise against trying to design an app that way in the first place. As maddy pointed out it's also likely considered to violate the HIGs. Even if you're not trying to do anything malicious and properly explain the feature in your app's description, that won't make Apple let it slide (I assume). Personally, as a user, I'd also find it annoying if the app did something the system already has a specific mechanic for in a different manner, at least in terms of app's coming to background and foreground.
I don't think it can be done without user interaction
The option is you can generate a push notification to tell the user to bring the application to foreground
When the operating system delivers push notification and the target application is not running in the foreground, it presents the notification.
If there is a notification alert and the user taps or clicks the action button (or moves the action slider), the application launches and calls a method to pass in the local-notification object or remote-notification payload.

How to detect if UIAccessibilityRequestGuidedAccessSession is active on device, iOS swift?

I know how to put app into Single app mode programmatically, provided that Autonomous single app mode persimmon is granted by MDM server to App.
This link have detail description about how to lock app in single app mode too.
Code to Apply single app mode as below -
UIAccessibilityRequestGuidedAccessSession(true){
success in
completionBlock(success)
}
My Question/Requirement is, detect if app is running in Autonomous single single app mode or UIAccessibilityRequestGuidedAccessSession is enabled, if it's enabled then only show alert to user and ask if he wish to disable Single App mode.
I Tried to detect using UIAccessibilityIsGuidedAccessEnabled() but it's of no use, as return value is always false.
You can use BOOL UIAccessibilityIsGuidedAccessEnabled(void); to get that information.
Source#AppleDocs
You could also try to add UIGuidedAccessRestrictionDelegate and then react to
func UIGuidedAccessRestrictionStateForIdentifier(_ restrictionIdentifier: String) -> UIGuidedAccessRestrictionState
Remember though, guided access needs to be enabled by the user (triple tap home button). Not from the settings!
So #Akaino answer is right, but UIAccessibilityIsGuidedAccessEnabled method wasn't working as expected due to i used to apply code below on didFinishLaunchingWithOptions hence it wasn't working properly
UIAccessibilityRequestGuidedAccessSession(true){
success in
completionBlock(success)
}
When i applied same code above on viewDidLoad() method, UIAccessibilityIsGuidedAccessEnabled is working as expected.

Unable to interact with system dialog (passcode) using XCUITest

My application opens a view controller which requires the user to enter their passcode to continue. I am writing UI automation and have used the record option to attempt to discover the element name that I need to interact with. The recorder tells me to use the following:
app.secureTextFields["Passcode field"].tap()
However when I run the test the secureTextField is not found. When I print out the elements in app there is no evidence of the view controller that is currently in the foreground and only the elements of my own application (which is in the background currently) are listed. I have also attempted to use addUIInterruptionMonitor (suggestions found here). The test fails on app.tap() stating that the application for target application is not in the foreground.
Is anyone aware of how to access elements of a system dialog currently in the foreground using XCTTest?

How to check application state under swift UI Test

Some background
I am currently writing a UI Test for a settings pane, and click on buttons to enable certain permissions such as push notifications and location services.
However, if the alert for the permission has been displayed before (regardless of the user allowing or denying access to the permission), the alert will not display again, and will just take the user to the settings app. Unfortunately, these settings do not reset, meaning the first time I run the UI tests, alerts will show; and on all subsequent UI test runs, the buttons will take me to the settings app unless I reset the device before the tests begin.
My issue
Thus, my test needs to know if the app went into the background, and attempt to foreground it to continue the testing:
if app.state == background {
foregroundApp()
}
// continue with other tests
Is there any way to determine if the app is in the background?
What I tried
I researched methods to determine the state of the application (running/background/etc) from a UI test, and was not able to find much. I tried to check whether certain elements exist:
if (app.navigationBars.element.exists) ...
but this gives me runtime errors[1] if the user is taken to the settings page because the app under test is in the background, and the test cannot lookup the navigationBars (or other elements).
I tried using some of the methods from Facebook's private headers for XCUIApplication() and XCUIElement().
XCUIApplication().state always returns 3 no matter what state the app is currently in, and any attempts to call XCUIApplication().resolve() to foreground the app give me the same errors as before[1]
I tried to rewrite the logic to foreground the app before resuming the tests, but methods such as XCUIApplication().launch() kill the app before restarting, which I cannot do. Only siri service seems to work for me, but I cannot access the siri service through the corporate proxy, and modifying proxy permissions is not possible.
Is there any other way to check the app state?
Errors
[1] This error is printed every time I try to do something involving state. I do not call snapshotView anywhere, and thus the suggestion to use afterScreenUpdates is useless.
Failure to get snapshot within 15.0s
Cannot snapshot view (<UIKeyboardImpl: 0x7febcc75d000; frame = (0 0;
414 226); layer = <CALayer: 0x608000625720>>) with
afterScreenUpdates:NO, because the view is not in a window. Use
afterScreenUpdates:YES.`
tl;dr
I need to check whether the app I am UI testing has entered the background (i.e. user pressed the home button). Checking for existence of particular elements such as navigation bars doesn't work, neither do most methods from Facebook's private headers for XCUIApplication/XCUIElement. Foregrounding the app also causes issues, and relaunching the app is not an option; neither is siri service.
You can do this in Swift 4, using XCUIApplication.state, which will give you information about the state of the app - whether it's in the foreground or background etc. however, it's not possible to find this information in Swift 3 and below. Essentially, UI testing in Swift 3 doesn't support leaving the app.

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