Swfit中视图跳转

1.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)

2.从当前视图跳转到下一视图

var?vc?=?AnswerViewController()
self.presentViewController(vc,?animated:?true,?completion:?nil)

3.通过dismissViewControllerAnimated(completion:)返回上一个视图

self.dismissViewControllerAnimated(true,245);">4.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
}
 
5.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)
 
6.通过 func popViewControllerAnimated() -> UIViewController! 弹出最上面的视图，并返回下一个视图控制器
 
7.通过func popToViewController(animated:) -> AnyObject[]!返回到navigationController视图堆栈中指定的某一个视