使用任何手势关闭 UIPopoverPresentationController，而不仅仅是点击

Question

所以我有一个显示一些内容的简单 UIPopoverPresentationController。

用户可以通过点击屏幕上的任意位置来关闭它（默认弹出行为）。

我希望在用户在屏幕上进行任何类型的点击或手势时关闭弹出窗口。最好是拖动手势。

知道这是否可行吗？以及如何？

Answer 1

尝试使用touchesBegan:withEvent方法

override func touchesBegan(touches: Set<UITouch>, withEvent event: UIEvent?) {
        if let touch = touches.first {
            if touch.view == self.view {
                self.dismiss()
            } else {
                return
            }

        }
    }

Answer 2

VC 是弹出窗口中显示的视图。在 presentViewController:animated:completion: 块

[self presentViewController:vc animated:YES completion:^{
UIView *v1 = vc.view.superview.superview.superview;

            for (UIView* vx in v1.subviews) {
                Class dimmingViewClass =  NSClassFromString(@"UIDimmingView");
                if ([vx isKindOfClass:[dimmingViewClass class]])
                {
                    UIPanGestureRecognizer* pan = [[UIPanGestureRecognizer alloc] initWithTarget:self action:@selector(closePopoverOnSwipe)];
                    [vx addGestureRecognizer:pan];
                }
            }
}];

您有一个 UIDimmingView，其中包含将关闭的点击手势。只需添加即可。我正在使用 Class dimmingViewClass = NSClassFromString(@"UIDimmingView"); 来避免直接使用未记录的 API。我还没有尝试将此 hack 发送给苹果，但下周会尝试。我希望它会过去。但我对此进行了测试，它确实调用了我的选择器。

Answer 3

我使用自定义视图解决了这个问题：

typealias Handler = (() -> Void)?


final class InteractionView: UIView {

    var dismissHandler: Handler = nil

    override func hitTest(_ point: CGPoint, with event: UIEvent?) -> UIView? {
        return self
    }

    override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
        self.dismissHandler?()
    }
}

在 viewDidAppear 中配置这个视图并添加到 popover containerView：

fileprivate func configureInteractionView() {
    let interactionView = InteractionView(frame: self.view.bounds)
    self.popoverPresentationController?.containerView?.addSubview(interactionView)
    interactionView.backgroundColor = .clear
    interactionView.isUserInteractionEnabled = true
    interactionView.dismissHandler = { [weak self] in
        self?.hide()
    }
}

fileprivate func hide() {
    self.dismiss(animated: true, completion: nil)
}

Answer 4

我对这个问题的解决方案。例如，如果您创建一个名为 MyPopoverViewController 的 class UIViewController 来呈现 PopViewController。然后在 viewDidLoad() 或 viewWillAppear(_ animated:) 方法中添加两个 GestureRecognizer 如下：

protocal MyPopoverControllerDelegate {
    func shouldDismissPopover()
}

class MyPopoverViewController : UIViewController {
    override func viewDidAppear(_ animated: Bool) {
    super.viewDidAppear(animated)

    // back trace to root view, if it is a UIWindows, add PanGestureRecognizer
    // and LongPressGestureRecognizer, to dismiss this PopoverViewController
    for c in sequence(first: self.view, next: { [=10=].superview}) {
        if let w = c as? UIWindow {

            let panGestureRecognizer = UIPanGestureRecognizer(target: self, action: #selector(dismissPopover(gesture:)))
            w.addGestureRecognizer(panGestureRecognizer)

            let longTapGestureRecognizer = UILongPressGestureRecognizer(target: self, action: #selector(dismissPopover(gesture:)))
            w.addGestureRecognizer(longTapGestureRecognizer)
        }
    }

    @objc private func dismissPopover(gesture: UIGestureRecognizer) {
        delegate?.shouldDismissPopover()
    }
}

然后在这个 PopOverViewController 呈现的主要 ViewController 中实现协议方法。

extension YourMainViewController: MyPopoverControllerDelegate {
    func shouldDismissPopover() {
        self.presentedViewController?.dismiss(animated: true, completion: nil)
    }
}

使用任何手势关闭 UIPopoverPresentationController，而不仅仅是点击

Dismiss UIPopoverPresentationController with any gesture, not just tap

uigesturerecognizer

ios

swift