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。
