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
对于那些使用@akashivsky的答案来实例化UIViewController的人来说，有一个异常:
致命错误:类使用了未实现的初始化式'init(coder:)'
快速提示:
手动实现所需的初始化?(coder aDecoder: NSCoder)在你的目标UIViewController，你试图实例化
required init?(coder aDecoder: NSCoder) {
super.init(coder: aDecoder)
}
如果你需要更多的描述，请参考我在这里的回答
其他回答
if let destinationVC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "DestinationVC") as? DestinationVC{
let nav = self.navigationController
//presenting
nav?.present(destinationVC, animated: true, completion: {
})
//push
nav?.pushViewController(destinationVC, animated: true)
}
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];
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)
}
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)
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
