swift - 在程序化触发的segue中传递对象？

Question

我已经阅读了很多资料，试图弄清楚这一点。究竟如何在以编程方式创建的 segue 中传递对象？我试图使 tableviewcell 成为弹出窗口启动的原因，并且从所有阅读中来看，设置 anchor 的唯一方法就是像下面这样。如何在这样的情况下传递对象？

当前代码:

func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {

    var deckObject: AnyObject = frc.objectAtIndexPath(indexPath) as! DeckResults
    let cell: victorycellcontroller = winRecord.cellForRowAtIndexPath(indexPath) as! victorycellcontroller
    let storyboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
    var playersNotesController: PlayerNotesViewController = storyboard.instantiateViewControllerWithIdentifier("StatWidgetPlayersNotes") as! PlayerNotesViewController
    playersNotesController.modalPresentationStyle = UIModalPresentationStyle.Popover
    let popoverplayersNoteController = playersNotesController.popoverPresentationController
    popoverplayersNoteController?.permittedArrowDirections = .Any
    popoverplayersNoteController?.delegate = self
    popoverplayersNoteController?.sourceView = cell.centerpointforpopover
    presentViewController(playersNotesController, animated: true, completion: nil)

}

Answer 1

让我们假设您要传递的对象是 itemToPass。

现在，在 PlayerNotesViewController 类中创建一个对象。

var yourObject: AnyObject!

现在，更新您的代码，如下所示:

var playersNotesController: PlayerNotesViewController = storyboard.instantiateViewControllerWithIdentifier("StatWidgetPlayersNotes") as! PlayerNotesViewController
playersNotesController.yourObject = itemToPass
playersNotesController.modalPresentationStyle = UIModalPresentationStyle.Popover

您会发现 itemToPass 中的值在 PlayerNotesViewController 中无法作为 yourObject 使用。