Question

我在iOS Swift应用程序中实现了一些UIPreviewActions。但是，这些操作执行segue事务，这些事务被忽略。我怎么能做到这一点？

有一个类似的问题，但在Objective-C中：UIPreviewAction to Mail From Peek

override func previewActionItems() -> [UIPreviewActionItem] {
    let performSegueAction = UIPreviewAction(title: "Go To View", style: .Default) { (action, viewController) -> Void in
    self.performSegueWithIdentifier("segueID", sender: self)
    }
    return [performSegueAction]
}

Answer 1

很抱歉迟到的回复，但我最近遇到了这个，所以发布答案以防其他人认为它有用......

在调用UIPreviewAction中的处理程序时，视图控制器已从层次结构中删除，因此执行segues无效。相反，你应该做类似的事情：

class PreviewViewController: UIViewController {

var performAction: (() -> Void)?

override func previewActionItems() -> [UIPreviewActionItem] {
    let performSegueAction = UIPreviewAction(title: "Go To View", style: .Default) { (action, viewController) -> Void in
        self.performAction?()
    }
    return [performSegueAction]
}

在父控制器中（从中启动预览）：

class ParentViewController: UIViewController {

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let vc = segue.destination as? PreviewViewController {
        vc.handleAction = {
            self.performSegueWithIdentifier("segueID", sender: self)
        }
    }
    inject(toController: segue.destination)
}

3DTouch UIPreviewAction：performseguewithidentifier无法正常工作

1 个答案: