Programmatically sending an app to background

非 Y 不嫁゛ 提交于 2019-12-17 19:36:12

问题


Is there a way to send the application to background? Similarly to how you can call XCUIApplication.terminate(), I have some UI Elements to test on applicationDidBecomeActive(_:). Does anyone know if this is at all possible?


回答1:


I would recommend checking out XCUIDevice. Here is how you might press the home button and then relaunch the application

func testExample() {

    // Has a nav bar.
    XCTAssert(XCUIApplication().navigationBars.element.exists)

    XCUIDevice().press(XCUIDeviceButton.home)
    // Before Swift 3: XCUIDevice().pressButton(XCUIDeviceButton.Home)
    XCUIApplication().launch()

    // Navigationbar still there on second launch.
    XCTAssert(XCUIApplication().navigationBars.element.exists)
}



回答2:


I just tried UIApplication.sharedApplication().performSelector("suspend") successfully.

dispatch_after(2, dispatch_get_main_queue(), {       
    // suspend the app after two seconds
    UIApplication.sharedApplication().performSelector("suspend")
})

// Swift 4 version
UIApplication.shared.perform(Selector("suspend"))



回答3:


In Xcode 9.0, Apple introduced XCUIApplication.activate(). activate() will launch the application if necessary, but will not terminate it if it is already running. Thus:

func testExample() {
    // Background the app
    XCUIDevice().press(.home)
    // Reactivate the app
    XCUIApplication().launch()
}



回答4:


In my case I wanted to background the app and open it where I previously left it. To background the app:

XCUIDevice.shared.press(.home)

To re-open the app where I left it:

XCUIApplication().activate()

Re-launching the app using XCUIApplication().launch() would launch the app from new.

I am using Swift 4.2



来源:https://stackoverflow.com/questions/33652978/programmatically-sending-an-app-to-background

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