Question

我想通过使用WKTapGestureRecognizer在watchOS上检测类似于“touchesBegan”和“touchesEnded”的状态。但是下面的代码总是返回状态3（触摸结束）。

- (IBAction)handleGesture:(WKTapGestureRecognizer*)gestureRecognizer{
NSLog(@":%ld",(long)gestureRecognizer.state);}

你能告诉我如何检测1号和2号国家。

Answer 1

正如official documentation中所述，WKTapGestureRecognizer是一个离散手势识别器，这意味着只有当它完成识别它时才会触发它，即在用户快速点击然后加注之后他的手指。

要从识别器接收多个状态更改，您必须使用连续识别器，例如WKLongPressGestureRecognizer（PanGesture并不是很有效）：如果设置的值非常低minimumPressDuration - 例如0.1秒 - 你可以欺骗它的行为类似于点击识别器，如下所示：

@IBAction func onRecognizerStateChange(_ recognizer: WKLongTapGestureRecognizer) {
    switch recognizer.state {
    case .began:
        print("User just touched the screen, finger is still resting")
    case .ended:
        print("User has raised his finger")
    default:
        break
    }
}

请记住，无论allowableMovement属性值如何，在识别器转换到Begin状态后，仍然允许用户拖动手指，如果您正在尝试，这可能是不可取的检测简单的水龙头。要解决此问题，您可以在识别器切换到Begin状态时存储初始触摸位置，然后检查每次Changed状态回调时触摸是否移动得太远，如下例所示：

private var touchLocation = CGPoint.zero

@IBAction func onRecognizerStateChange(_ recognizer: WKLongTapGestureRecognizer) {
    switch recognizer.state {
    case .began:
        touchLocation = recognizer.locationInObject()
        print("User just touched the screen at \(touchLocation)")
    case .changed:
        let location = recognizer.locationInObject()
        let distance = sqrt(pow(touchLocation.x - location.x, 2.0) 
            + pow(touchLocation.y - location.y, 2.0))
        if distance >= 10 {
              print("User traveled too far from \(touchLocation)")
              // invalidate the gesture recognizer
              recognizer.isEnabled = false
              recognizer.isEnabled = true
        }
    case .ended:
        print("User successfully completed a tap-like gesture")
    default:
        break
    }
}

在watchOS上进行触摸检测

1 个答案: