Swfit中视图跳转

.跳转到任一UIViewController

  var sb = UIStoryboard(name: "Main", bundle:nil)
  var vc = sb.instantiateViewControllerWithIdentifier("ChooseViewController") as ChooseViewController
  self.presentViewController(vc, animated:true, completion:nil) .从当前视图跳转到下一视图   var vc = AnswerViewController()
  self.presentViewController(vc, animated: true, completion: nil) .通过dismissViewControllerAnimated(completion:)返回上一个视图   self.dismissViewControllerAnimated(true, completion:nil) .Modal Segue to channel Controller
通过在storyboard设计视图中,选择一个按钮,右键拖动到另一个视图,即可建立动作跳转,但需要重载func prepareForSegue(segue: UIStoryboardSegue!, sender: AnyObject!)方法,如下:   override func prepareForSegue(segue: UIStoryboardSegue!, sender: AnyObject!) {
   var channelC:ChannelController=segue.destinationViewController as ChannelController
  channelC.delegate=self
   channelC.channelData=self.channelData
  } .通过navigationController.pushViewController(animated:)方法   var webView=WebViewController()
  webView.detailID=data.newsID
  //取导航控制器,添加subView
self.navigationController.pushViewController(webView,animated:true) .通过 func popViewControllerAnimated() -> UIViewController! 弹出最上面的视图,并返回下一个视图控制器 .通过func popToViewController(animated:) -> AnyObject[]!返回到navigationController视图堆栈中指定的某一个视图
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