Question

我有一个UITableViewController显示为弹出窗口，如下所示

    popoverViewController = self.storyboard?.instantiateViewController(withIdentifier: "ItemsTableView") as? ItemsTableViewController
    popoverViewController?.itemeSelectionDelegate = self
    popoverViewController?.modalPresentationStyle = .popover

    let popoverPresentationViewController = popoverViewController?.popoverPresentationController
    let itemCell  = cell as! ItemContentCell
    popoverPresentationViewController?.permittedArrowDirections = UIPopoverArrowDirection.any
    popoverPresentationViewController?.sourceView = cell
    popoverPresentationViewController?.sourceRect = itemCell.itemName.bounds

    present(popoverViewController!, animated: true, completion: nil)

popoverViewController是一个成员var，我用它来关闭当从popoverViewController中选择一个项目时，我正在调用一个自定义委托方法，我从中解雇它使用“Dismiss”方法但它有时候工作有时它不会。

 func itemSelected(item: Item) {
    print("selected item")
    popoverViewController?.dismiss(animated: false, completion: nil)
}

还有其他方法可以解雇popover吗？

Answer 1

您必须实施UIPopoverPresentationControllerDelegate方法public func popoverPresentationControllerShouldDismissPopover(_ popoverPresentationController: UIPopoverPresentationController) -> Bool。这是演示实现。

    extension YouViewController: UIPopoverPresentationControllerDelegate {

        func popoverPresentationControllerShouldDismissPopover(_ popoverPresentationController: UIPopoverPresentationController) -> Bool {
            return true
        }
    }

不能解雇Popover UITableViewController

1 个答案: