Ios 具有表视图的自定义交互式转换

Ios 具有表视图的自定义交互式转换,ios,swift,animation,custom-transition,Ios,Swift,Animation,Custom Transition,我创建了自定义推/弹出交互转换,但我不知道如何同时处理此转换和表视图滚动 对于控制柄自定义弹出动画,我添加平移手势 let pan = UIPanGestureRecognizer( target: self, action: #selector(didPan)) pan.delegate = self view.addGestureRecognizer(pan) 接下来我来拿这个锅 if recognizer.state ==

我创建了自定义推/弹出交互转换,但我不知道如何同时处理此转换和表视图滚动

对于控制柄自定义弹出动画,我添加平移手势

let pan = UIPanGestureRecognizer(
                target: self,
                action: #selector(didPan))
pan.delegate = self
view.addGestureRecognizer(pan)
接下来我来拿这个锅

if recognizer.state == .began {
    animator.interactive = true
    navigationController!.popViewController(animated: true)
}

animator.handlePan(recognizer: recognizer)
然后,内部动画师遵循
uipercentdriventeractivetransition
UIViewControllerAnimatedTransitioning
I转换动画。 我尝试通过平移手势将导航从VC_2弹出到VC_1

问题是添加平移手势后,表格视图没有滚动。我想这是因为有两个手势识别器(我自己的和UIKit表格视图的识别器)。事实上,当表格向上滚动时,我希望我自己的pan以交互方式弹出我的VC_2

唯一的想法是

func gestureRecognizer(
        _ gestureRecognizer: UIGestureRecognizer,
        shouldRecognizeSimultaneouslyWith otherGestureRecognizer: UIGestureRecognizer) -> Bool {
    return tableView.contentOffset.y <= 0
}
func手势识别器(
_手势识别器:UIGestureRecognitor,
应与其他手势识别器同时识别:UIGestureRecognitor)->Bool{
返回tableView.contentOffset.y