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
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];
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。
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)
}
斯威夫特4:
let storyboard = UIStoryboard(name: "Main", bundle: nil)
let yourVC: YourVC = storyboard.instantiateViewController(withIdentifier: "YourVC") as! YourVC
