Swift 3将变量传递给PopupViewController

问题描述 投票:0回答:2

我使用accessoryButtonTappedForRowWith呈现我的viewController:

let storyboard : UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
        let popoverContent = storyboard.instantiateViewController(withIdentifier: "Test")

        popoverContent.modalPresentationStyle = .popover

        if let popover = popoverContent.popoverPresentationController {
            let viewForSource = self.createCharacterView as UIView
            popover.sourceView = viewForSource

            // the position of the popover where it's showed
            popover.sourceRect = viewForSource.bounds

            // the size you want to display
            popoverContent.preferredContentSize = CGSize(width: 200,height: 200)
            popover.delegate = self
        }

        self.present(popoverContent, animated: true, completion: nil)

弹出工作并且看起来很好,只是我呈现的viewController需要tableViewCell的标签。我能够得到变量,但我不确定如何将变量传递给popover因为没有调用segue我不认为我能够使用prepare(for segue: UIStoryboardSegue, sender: Any?)

ios swift3 popover
2个回答
0
投票

这太简单了:

popoverContent.variableName = passingVariable

0
投票

您可能需要将popOvercontent转换为特定的viewController

(popoverContent as? MyViewController).myVar = value
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