Make XCUIElement perform 3D touch for Automate UITest?

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轮回少年
轮回少年 2021-01-04 18:50

I\'m creating Automate UITest test case where I\'d like to test the scenario when users make 3D Touch interaction with an element, then shows them Peek and Pop view.

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  •  刺人心
    刺人心 (楼主)
    2021-01-04 19:14

    I was able to execute a force press / 3D Touch on an app icon on an iPhone 7 running iOS 10.3. It was not possible on 10.2.

    Objective-C

    First you have to declare the following or import this header

    typedef void (^CDUnknownBlockType)(void);
    
    @interface XCEventGenerator : NSObject
    
    + (id)sharedGenerator;
    
    // iOS 10.3 specific
    - (double)forcePressAtPoint:(struct CGPoint)arg1 orientation:(long long)arg2 handler:(CDUnknownBlockType)arg3;
    
    @end
    

    And then execute the force press

    XCUIElement *element = ...; // Get your element
    XCUICoordinate *coord = [mapsIcon coordinateWithNormalizedOffset:CGVectorMake(0.5, 0.5)];
    
    [[XCEventGenerator sharedGenerator] forcePressAtPoint:coord.screenPoint orientation:0 handler:^{}]; // handler cannot be nil
    

    Here I execute a force press on the Maps icons.

    Swift (not tested)

    For Swift you have to declare/import the same interface/header as above and then execute the force press like this

    let el = ...; // Get your element
    let coord = el.coordinate(withNormalizedOffset: CGVector(dx: 0.5, dy: 0.5) )
    let eventGenerator: XCEventGenerator = XCEventGenerator.sharedGenerator() as! XCEventGenerator
    
    eventGenerator.forcePress(at: coord.screenPoint, orientation: 0) { }
    

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