Question

我在主viewController上有一个按钮，其默认值nil由dropDown pod关联。在同一个viewController上还有一个容器视图。

在首次加载期间，我从共享首选项中获取变量的默认值，并通过performSegue将该值传递给容器视图。

 override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
   if(segue.identifier == "dataToContainerView"){
    DispatchQueue.main.async {

   var secondVC = segue.destination as! secondViewController //container viewController
        secondVC.variable = self.variable  
    }
  }
}

现在我需要通过用户选择dropdown按钮再次传递同一个变量的值。

 dropDown.selectionAction = { [unowned self] (index, item) in
        self.button.setTitle(item, for: UIControlState())
        self.variable = item
        print(item)
        self.performSegue(withIdentifier: "dataToContainerView", sender: nil)
  //performing segue to resend the new value of the variable.
    }

上述代码正常执行到print(item)。但是我在performSegue上收到以下错误。

Terminating app due to uncaught exception 'NSInternalInconsistencyException', reason: 'There are unexpected subviews in the container view. Perhaps the embed segue has already fired once or a subview was added programmatically?

我应该如何在dropDown pod的帮助下将值传递给容器视图第二次覆盖第一个值？

更新： - 我需要变量值，以便我可以将它传递给容器viewController上的json解析器。容器viewController上的代码重新执行。

Answer 1

您需要保存对嵌入式控制器的引用，以便以后再次更新。在第一个segue中执行此操作：

// Declare a local variable in your parent container:
var secondVC: secondViewController!

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if(segue.identifier == "dataToContainerView"){
        DispatchQueue.main.async {
            self.secondVC = segue.destination as! secondViewController 
            //container viewController
            self.secondVC.variable = self.variable  
        }
    }
}

然后，当您需要更新变量时，您可以直接引用它：

self.secondVC.variable = self.variable
self.secondVC.viewDidLoad()

将相同的变量传递给具有不同值的容器视图

1 个答案: