Question

感谢Austin解决了我的问题，现在更新了正确的代码：（以下代码描述了问题：）

import Cocoa

class MyViewController1 : NSViewController, myProtocol {

// Identifier MyViewController2ID is set in storyboard for class MyViewController2
lazy var myViewController2: NSViewController = {
    return self.storyboard!.instantiateControllerWithIdentifier("MyViewController2ID")
        as! NSViewController
}()

// ViewController conforms to protocol
func myFunction() {
    // do something
}

// ... some action
// presenting new viewcontroller as sheet
// new controller is presented and is dismissed
// when self.dismissController(self) is called
// in MyViewController2

self.presentViewControllerAsSheet(self.myViewController2)
}

protocol myProtocol : class {
func myFunction()
}

// Identifier MyViewController2ID for class 
// MyViewController2 is set in storyboard
class MyViewController2 : NSViewController {

weak var delegate:myProtocol?

override func viewDidLoad() {
    super.viewDidLoad()
    ...
     if let pvc = self.presentingViewController as? MyViewController1 {
        self.delegate = pvc
    }
 }
// function is activated
func someFunction() {
    self.delegate?.myFunction()
    self.dismissController(self)
   }

 }

MyViewController2通过调用：

呈现

self.presentViewControllerAsSheet（self.myViewController2）

在MyViewController1中

并在someFunction（）中调用self.dismisscontroller（self）在MyViewController2中解除..

问题是委托是无...我不知道如何传递对委托函数的引用？

当调用另一个viewcontroller bye segue时，在segue中设置对委托函数的引用。没有segue为委托函数提供参考？

任何？

Answer 1

在MyViewController2的viewDidLoad()中，获取presentingViewController并将其设置为委托。

// viewDidLoad - MyViewController2
if let pvc = self.presentingViewController as? MyViewController1 {
    self.delegate = pvc
}

委托功能，如何传递参考？

1 个答案: