Possible to bring the app from background to foreground?

Question

When running an XCT UI test it is possible to put the application under test in the background with:

XCUIDevice().pressButton(XCUIDeviceButton.Home)

It it possible in some way to bring the app back to foreground (active state) without relaunching the application?

Carola

Swift 5 syntax: XCUIDevice.shared.press(XCUIDevice.Button.home) – Batish

Answer 1

Update for Xcode 9: Starting in Xcode 9, you can now simply call activate() on any XCUIApplication.

let myApp = XCUIApplication()
myApp.activate() // bring to foreground

https://developer.apple.com/documentation/xctest/xcuiapplication/2873317-activate

---

Yes, it is. But, you'll need XCUIElement's private headers (which are available via header dump from Facebook here). In order to foreground the app, you need to call resolve which I believe resolves the element's query (which for applications means foregrounding the app).

For Swift, you'll have to import the XCUIElement.h into your bridging header. For Objective-C you'll just need to import XCUIElement.h.

With the app backgrounded:

Swift:

XCUIApplication().resolve()

Objective-C

[[XCUIApplication new] resolve];

If this is the only functionality you need, you could just write a quick ObjC category.

@interface XCUIElement (Tests)
- (void) resolve;
@end

If you need to launch / resolve another app. Facebook has an example of that here by going through the Springboard.

Bill

I'm sure this method also works, but it's a lot of workaround for what you can simply accomplish by using XCUIApplication().launch(). – Formula

When I wrote this -- calling launch() killed the app and restarted it. As far as I know, you still need to call resolve() if you need to foreground the app without relaunching it. – Bill

I think you're right. I was only thinking about launching the app (not without killing the app first). – Formula

Is this still working for you guys? I feel @brandenbyers solution might just be better. – Kisser

How do you do this part?: you'll have to import the XCUIElement.h into your bridging header – Branchia

Yes @rfodge, you'll need to import the category file into your bridging header. As of Xcode 9, this is no longer necessary, though, as there is a new public API to accomplish this task. – Bill

Yeah i saw that is what is needed to do but how exactly do you do that? Sorry i am not familiar with where i even find the bridging header, or what i would need to put. If you could let me know or add to above, that would be awesome! Thamks – Branchia

If you add any objective-c class file to your test target (via the File -> New File, not dragging in), Xcode will give you an option to create a bridging header automatically. Just replace the contents of the created file with the XCUIElement interface. – Bill

The .activate() doesn't seem to be working in Xcode 10.2. It always fails for me. – Smetana

Answer 2

As of Xcode 8.3 and iOS 10.3, you can accomplish this with Siri:

XCUIDevice.shared().press(XCUIDeviceButton.home)
XCUIDevice.shared().siriService.activate(voiceRecognitionText: "Open {appName}")

Include @available(iOS 10.3, *) at the top of your test suite file and you should be good to go!

Halliburton

It is important to note that to get this to work on the iOS Simulator you need to enable Siri in the Settings app (off by default). I'm not sure if there is a way to turn this on via a command line argument or anything unfortunately. – Litman

Good point. Siri must be enabled for this to work whether a physical device or simulator. Luckily, siri only need be enabled once on each simulated device. From my experience the setting remains between relaunches of simulator. I haven't found a way to enable siri on simulators via the command line. – Halliburton

Answer 3

This is what I have in my XCUITest and it works like a charm (xcode 10.1 and test device is iPhone X 11.0)

func testWhatever() {

// You test steps go here until you need the background foreground to run

XCUIDevice.shared.press(XCUIDevice.Button.home) // To background the app XCUIApplication().activate() // To bring the app back

// You test continues after background foreground has been done. }

Dagall

Answer 4

If somebody needs just move app back from background i have written (based on answer above) category that really works(great thanks to pointing to FB git)

@implementation XCUIApplication(SpringBoard)

+ (instancetype)springBoard
{
    XCUIApplication * springboard  = [[XCUIApplication alloc] performSelector:@selector(initPrivateWithPath:bundleID:)
                                                                   withObject:nil
                                                                   withObject:@"com.apple.springboard"];




    [springboard performSelector:@selector(resolve) ];
    return springboard;
}

- (void)tapApplicationWithIdentifier:(NSString *)identifier
{
    XCUIElement *appElement = [[self descendantsMatchingType:XCUIElementTypeAny]
                           elementMatchingPredicate:[NSPredicate predicateWithFormat:@"identifier = %@", identifier]
                           ];
    [appElement tap];
}
@end

Metacarpal

Answer 5

For Swift, you need to declare the XCUIApplication private methods interface in Bridging-Header.h like this:

@interface XCUIApplication (Private)
- (id)initPrivateWithPath:(NSString *)path bundleID:(NSString *)bundleID;
- (void)resolve;
@end

Then call resolve() in your test cases to bring the app back:

XCUIApplication().resolve()

Beghard

Where do you find the Brigding-Header.h file? – Branchia

It's automatically created when you add an objc file to your Swift target. If you don't already have one, the easiest thing to do is to just create an empty objc file and add it to your target. Xcode will ask if you want a bridging header -- say yes. Then delete the empty objc file you created. The bridging header will remain. Otherwise you can create a header file and set its path in Project->Build Settings->Objective-C Bridging Header. – Bill

Answer 6

Since Xcode 13 we got several errors that the app was not in foreground state after returning to the app.

applying this code to our "goToSpringboardAndBack()" works

XCUIDevice.shared.press(XCUIDevice.Button.home)
if XCUIApplication().wait(for: .runningBackground, timeout: 5.0) {
   XCUIApplication().activate()
}

_ = XCUIApplication().wait(for: .runningForeground, timeout: 5.0)
´´´

Hightension