问题

我开始看看Swift编程语言,不知为何我不能正确地从特定的UIStoryboard输入一个UIViewController的初始化。

在Objective-C中,我简单地写:

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"StoryboardName" bundle:nil];
UIViewController *viewController = [storyboard instantiateViewControllerWithIdentifier:@"ViewControllerID"];
[self presentViewController:viewController animated:YES completion:nil];

有人能帮助我如何在斯威夫特上实现这一点吗?


当前回答

斯威夫特5

let vc = self.storyboard!.instantiateViewController(withIdentifier: "CVIdentifier")
self.present(vc, animated: true, completion: nil)

其他回答

akashivsky的答案很好!但是,如果你从呈现的视图控制器返回时遇到一些麻烦,这个替代方法会很有用。这对我很管用!

迅速:

let storyboard = UIStoryboard(name: "MyStoryboardName", bundle: nil)
let vc = storyboard.instantiateViewControllerWithIdentifier("someViewController") as! UIViewController
// Alternative way to present the new view controller
self.navigationController?.showViewController(vc, sender: nil)

Obj - c:

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"MyStoryboardName" bundle:nil];
UIViewController *vc = [storyboard instantiateViewControllerWithIdentifier:@"someViewController"];
[self.navigationController showViewController:vc sender:nil];

斯威夫特3

let settingStoryboard : UIStoryboard = UIStoryboard(name: "SettingViewController", bundle: nil)
let settingVC = settingStoryboard.instantiateViewController(withIdentifier: "SettingViewController") as! SettingViewController
self.present(settingVC, animated: true, completion: {

})

Swift 4.2更新的代码是

let storyboard = UIStoryboard(name: "StoryboardNameHere", bundle: nil)
let controller = storyboard.instantiateViewController(withIdentifier: "ViewControllerNameHere")
self.present(controller, animated: true, completion: nil)

斯威夫特5

let vc = self.storyboard!.instantiateViewController(withIdentifier: "CVIdentifier")
self.present(vc, animated: true, completion: nil)

我创建了一个库,可以用更好的语法更容易地处理这个问题:

https://github.com/Jasperav/Storyboardable

只需改变Storyboard.swift,让ViewControllers符合Storyboardable。