在iOS6中,我注意到新的容器视图,但不太确定如何从包含视图访问它的控制器。

场景:

我想从包含容器视图的视图控制器中访问Alert视图控制器中的标签。

它们之间有一个segue,我能用吗?


当前回答

你可以这样写

- (IBAction)showDetail:(UIButton *)sender {  
            DetailViewController *detailVc = [self.childViewControllers firstObject];  
        detailVc.lable.text = sender.titleLabel.text;  
    }  
}

其他回答

Swift编程

你可以这样写

var containerViewController: ExampleViewController?
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
    // you can set this name in 'segue.embed' in storyboard
    if segue.identifier == "checkinPopupIdentifierInStoryBoard" {
        let connectContainerViewController = segue.destinationViewController as ExampleViewController
        containerViewController = connectContainerViewController
    }
}

你可以这样写

- (IBAction)showDetail:(UIButton *)sender {  
            DetailViewController *detailVc = [self.childViewControllers firstObject];  
        detailVc.lable.text = sender.titleLabel.text;  
    }  
}

在视图控制器的类型上使用Swift的switch语句还有另一种方法:

override func prepare(for segue: UIStoryboardSegue, sender: Any?)
{
  switch segue.destination
  {
    case let aViewController as AViewController:
      self.aViewController = aViewController
    case let bViewController as BViewController:
      self.bViewController = bViewController
    default:
      return
  }
}

如果有人正在寻找Swift 3.0,

viewController1, viewController2等等都是可访问的。

let viewController1 : OneViewController!
let viewController2 : TwoViewController!

// Safety handling of optional String
if let identifier: String = segue.identifier {

    switch identifier {

    case "segueName1":
        viewController1 = segue.destination as! OneViewController
        break

    case "segueName2":
        viewController2 = segue.destination as! TwoViewController
        break

    // ... More cases can be inserted here ...

    default:
        // A new segue is added in the storyboard but not yet including in this switch
        print("A case missing for segue identifier: \(identifier)")
        break
    }

} else {
    // Either the segue or the identifier is inaccessible 
    print("WARNING: identifier in segue is not accessible")
}

是的,你可以使用segue来访问子视图控制器(及其视图和子视图)。使用Storyboard中的Attributes检查器给segue一个标识符(比如alertview_embed)。然后让父视图控制器(包含容器视图的控制器)实现如下方法:

- (void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
   NSString * segueName = segue.identifier;
   if ([segueName isEqualToString: @"alertview_embed"]) {
       AlertViewController * childViewController = (AlertViewController *) [segue destinationViewController];
       AlertView * alertView = childViewController.view;
       // do something with the AlertView's subviews here...
   }
}