xcode 快速访问容器视图子属性
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Access container view child properties swift
提问by Tom el Safadi
What I want to achieve:
我想要达到的目标：
User presses the button in the ViewController then, the color of the button placed in the container view should change its color to red.
用户按下 ViewController 中的按钮，然后放置在容器视图中的按钮的颜色应更改为红色。
How can I get access of the button placed in the container view, from the ViewController?
如何从 ViewController 访问放置在容器视图中的按钮？
回答by sunshinejr
Step by step:
一步步：
- Name the segue between your view controller and container view controller.
- Add a property to your view controller which will contain the container view controller.
- In your view controller implement a method
prepareForSegue(_:sender:). - In the method check if
segue.identifierequals the identifier you specified in step 1. - If true, then save the
segue.destinationViewControllerto your property from step 2. - Now you have the container view controller stored in your property so you can do customization from your class. You should have the view controller stored in
viewDidLoad()method already.
- 命名视图控制器和容器视图控制器之间的 segue。
- 向视图控制器添加一个属性，该属性将包含容器视图控制器。
- 在您的视图控制器中实现一个方法
prepareForSegue(_:sender:)。 - 在方法中检查是否
segue.identifier等于您在步骤 1 中指定的标识符。 - 如果为 true，则将
segue.destinationViewController第 2 步中的保存到您的属性中。 - 现在您已将容器视图控制器存储在您的属性中，以便您可以从您的类中进行自定义。您应该已经将视图控制器存储在
viewDidLoad()方法中。
Example:
例子：
var containerViewController: YourContainerViewControllerClass?
let containerSegueName = "testSegue"
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == containerSegueName {
containerViewController = segue.destinationViewController as? YourContainerViewControllerClass
}
}
回答by Andrey Gordeev
I recommend not to rely on segue.identifier, but rather test for destinationtype directly:
我建议不要依赖segue.identifier，而是destination直接测试类型：
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
super.prepare(for: segue, sender: sender)
if let vc = segue.destination as? YourViewController {
vc.someVariable = true
}
}
This way you avoid mistakes with a misspelled segue name.
这样你就可以避免拼写错误的 segue 名称的错误。
回答by Saranjith
Swift 4, Xcode 9.4.1
斯威夫特 4，Xcode 9.4.1
var contentViewController : UIContentViewController?
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
if segue.identifier == containerSegueName {
contentViewController = segue.destination as? UIContentViewController
}
}
回答by asdf
Swift 3 for macOS:
适用于 macOS 的 Swift 3：
// MARK: - Container View Controller
var containerViewController: ContainerViewController?
let containerSegueIdentifier = "Container Segue"
override func prepare(for segue: NSStoryboardSegue, sender: Any?) {
if segue.identifier == containerSegueIdentifier {
if let connectContainerViewController = segue.destinationController as? FormationViewController {
formationViewController = connectContainerViewController
}
}
}
Check identifier and controller class.
检查标识符和控制器类。
