Question

我一直在阅读相当多的试图解决这个问题。你如何在一个以编程方式创建的segue中传递一个对象？我正在尝试使tableviewcell成为启动popover的原因，并且从所有读取中设置锚点的唯一方法就是如下所示。你如何以这样的方式传递一个物体？

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PlayerNotesViewController的值不是yourObject {/ 1}}。

在程序触发的segue中传递对象？

1 个答案: