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
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
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
如果你有一个Viewcontroller没有使用任何storyboard/Xib，你可以像下面这样调用这个特定的VC:
let vcInstance : UIViewController = yourViewController()
self.present(vcInstance, 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)
}
这个答案是针对Swift 5.4和iOS 14.5 SDK进行的最新修订。
这只是新语法和稍微修改的api的问题。UIKit的底层功能并没有改变。对于绝大多数iOS SDK框架来说都是如此。
let storyboard = UIStoryboard(name: "myStoryboardName", bundle: nil)
let vc = storyboard.instantiateViewController(withIdentifier: "myVCID")
self.present(vc, animated: true)
确保在故事板中“故事板ID”下设置myVCID。
