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
// "Main" is name of .storybord file "
let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
// "MiniGameView" is the ID given to the ViewController in the interfacebuilder
// MiniGameViewController is the CLASS name of the ViewController.swift file acosiated to the ViewController
var setViewController = mainStoryboard.instantiateViewControllerWithIdentifier("MiniGameView") as MiniGameViewController
var rootViewController = self.window!.rootViewController
rootViewController?.presentViewController(setViewController, animated: false, completion: nil)
当我把它放在AppDelegate时，这工作得很好
其他回答
这个链接有两个实现:
迅速:
let viewController:UIViewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewControllerWithIdentifier("ViewController") as UIViewController
self.presentViewController(viewController, animated: false, completion: nil)
Objective - C
UIViewController *viewController = [[UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil] instantiateViewControllerWithIdentifier:@"ViewController"];
这个链接有在同一个故事板中初始化视图控制器的代码
/*
Helper to Switch the View based on StoryBoard
@param StoryBoard ID as String
*/
func switchToViewController(identifier: String) {
let viewController = self.storyboard?.instantiateViewControllerWithIdentifier(identifier) as! UIViewController
self.navigationController?.setViewControllers([viewController], animated: false)
}
我知道这是一个旧线程，但我认为目前的解决方案(使用硬编码的字符串标识符为给定的视图控制器)是非常容易出错。
我已经创建了一个构建时脚本(你可以在这里访问)，它将创建一个编译器安全的方式来访问和实例化给定项目中的所有故事板中的视图控制器。
例如，在Main中名为vc1的视图控制器。故事板将像这样实例化:
let vc: UIViewController = R.storyboard.Main.vc1^ // where the '^' character initialize the controller
斯威夫特4:
let storyboard = UIStoryboard(name: "Main", bundle: nil)
let yourVC: YourVC = storyboard.instantiateViewController(withIdentifier: "YourVC") as! YourVC
无论我怎么尝试，它都不适合我——没有错误，但我的屏幕上也没有新的视图控制器。不知道为什么，但是在超时函数中包装它最终使它工作:
DispatchQueue.main.asyncAfter(deadline: .now() + 0.0) {
let storyboard = UIStoryboard(name: "Main", bundle: nil)
let controller = storyboard.instantiateViewController(withIdentifier: "TabletViewController")
self.present(controller, animated: true, completion: nil)
}
我想建议一种更简洁的方法。当我们有多个故事板时，这将很有用
1.用你所有的故事板创建一个结构
struct Storyboard {
static let main = "Main"
static let login = "login"
static let profile = "profile"
static let home = "home"
}
2. 创建一个UIStoryboard扩展，如下所示
extension UIStoryboard {
@nonobjc class var main: UIStoryboard {
return UIStoryboard(name: Storyboard.main, bundle: nil)
}
@nonobjc class var journey: UIStoryboard {
return UIStoryboard(name: Storyboard.login, bundle: nil)
}
@nonobjc class var quiz: UIStoryboard {
return UIStoryboard(name: Storyboard.profile, bundle: nil)
}
@nonobjc class var home: UIStoryboard {
return UIStoryboard(name: Storyboard.home, bundle: nil)
}
}
将故事板标识符作为类名，并使用下面的代码进行实例化
let loginVc = UIStoryboard.login.instantiateViewController(withIdentifier: "\(LoginViewController.self)") as! LoginViewController
