为滚动视图上的两指滑动分配唯一行为

Question

我想为WKWebView上的两指滑动指定一种独特的行为。 提到这个站点，我编写了如下代码。 它起作用了，但是在Webview上滚动变得非常慢。

有没有更好的方法来避免缓慢滚动？

    let doubleSwipeGestureRecognizer = UISwipeGestureRecognizer.init(target: self, action: "doubleSwiped2")
    doubleSwipeGestureRecognizer.numberOfTouchesRequired = 2
    doubleSwipeGestureRecognizer.direction = UISwipeGestureRecognizerDirection.Up
    self.webView!.addGestureRecognizer(doubleSwipeGestureRecognizer)

    for gesture in self.webView!.scrollView.gestureRecognizers!{
        let gestureClass = gesture.classForCoder
        let gestureName = NSStringFromClass(gestureClass)
        print(gestureName)

        if gestureName.containsString("Swipe"){
            // do nothing
        } else {
            gesture.requireGestureRecognizerToFail(doubleSwipeGestureRecognizer)

        }
    }

Answer 1

也许此代码适合您的需求。

首先，将self设置为识别器的委托。

doubleSwipeGestureRecognizer.delegate = self

其次，在自类中编写UIGestureRecognizerDelegate的方法。

func gestureRecognizer(gestureRecognizer: UIGestureRecognizer, shouldBeRequiredToFailByGestureRecognizer otherGestureRecognizer: UIGestureRecognizer) -> Bool {
    if otherGestureRecognizer is UIPanGestureRecognizer && otherGestureRecognizer.numberOfTouches() == 2 {
        return true
    }

    return false
}

第三，删除“ for”句子。

for gesture in self.webView!.scrollView.gestureRecognizers! {
…
}