Swfit中视图跳转

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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, completion:nil)

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图

欢迎完善。。。。

Swfit中视图跳转

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原文地址：http://www.cnblogs.com/adjk/p/5261061.html