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
我使用这个助手:
struct Storyboard<T: UIViewController> {
static var storyboardName: String {
return String(describing: T.self)
}
static var viewController: T {
let storyboard = UIStoryboard(name: "Main", bundle: nil)
guard let vc = storyboard.instantiateViewController(withIdentifier: Self.storyboardName) as? T else {
fatalError("Could not get controller from Storyboard: \(Self.storyboardName)")
}
return vc
}
}
用法(故事板ID必须匹配UIViewController类名)
let myVC = Storyboard.viewController as MyViewController
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];
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
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
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)
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)
}
