ios UIStoryboard：获取活动故事板的正确方法是什么？
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原文地址: http://stackoverflow.com/questions/9853608/
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UIStoryboard: What's the Correct Way to Get the Active Storyboard?
提问by Chris Marshall
I am currently furiously digging through all the docs, and haven't quite found what I'm looking for. I suspect it is a real d'oh! answer.
我目前正在疯狂地翻阅所有文档，但还没有完全找到我要找的东西。我怀疑这是一个真正的 d'oh！回答。
I simply need to find the active storyboard in the main bundle, and want to know the best way to do this.
我只需要在主包中找到活动的故事板，并想知道这样做的最佳方法。
This is so that I can use the [UIStoryboard storyboardWithName:@"XXX" bundle:mainBundle]to extract the running storyboard.
这样我就可以使用[UIStoryboard storyboardWithName:@"XXX" bundle:mainBundle]来提取正在运行的故事板。
I know how to kludge it by switching on the idiom, but I feel that this is a...kludge.
我知道如何通过打开成语来混杂它，但我觉得这是一个......混杂。
What's a correct way of doing this?
这样做的正确方法是什么？
采纳答案by Chris Marshall
OK. As my comment above indicates, I found the answer to the (badly phrased question):
好的。正如我上面的评论所示，我找到了（措辞不当的问题）的答案：
I wanted to be able to get the main(not active) storyboard, as I'm not using multiple storyboards per incarnation. I'm using the standard model of 1 storyboard for iPhone, and 1 for iPad. I just wanted the cleanest way to get the storyboard, so that I could use it to generate a view controller.
我希望能够获得主要（非活动）故事板，因为我没有在每个化身中使用多个故事板。我正在使用 iPhone 的 1 个故事板和 iPad 的 1 个故事板的标准模型。我只想以最简洁的方式获取故事板，以便我可以使用它来生成视图控制器。
I found the answer in this post on Stack Overflow, and implemented it with the following code:
我在 Stack Overflow 上的这篇文章中找到了答案，并使用以下代码实现了它：
UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];
回答by olivaresF
In case you want to get the active storyboard for a viewController, there's a storyboard property. This is how I solved it, instead of making a new instance:
如果您想获取 viewController 的活动情节提要，则有一个情节提要属性。这就是我解决它的方法，而不是创建一个新实例：
LoginViewController *vc = [navController.storyboard instantiateViewControllerWithIdentifier:@"firstLaunch"];
[navController presentModalViewController:vc animated:YES];
In Swift you'd call:
在 Swift 中，您会调用：
let loginViewController = navigationController?.storyboard?.instantiateViewController(withIdentifier: "firstLaunch") as! LoginViewController
navigationController?.present(loginViewController, animated: true, completion: nil)
You could also be a lot safer by using guards against the navigation controller and the storyboard. I've used as!so as to guarantee that you're getting a LoginController.
通过使用对导航控制器和故事板的防护，您也可以更安全。我已经使用as!以便保证您获得 LoginController。
回答by nburk
In Swift, you'd use the following syntax:
在Swift 中，您将使用以下语法：
let storyboard = UIStoryboard(name: "Main", bundle: nil)
Note that passing nilto bundlewill make the call refer to your mainbundle automatically.
请注意，传递nil到bundle将使调用自动引用您的主包。
If you're in a view controllerthat you have on the Storyboard and want to instantiate the Storyboard from there directly, you can just do:
如果您在Storyboard 上的视图控制器中并想直接从那里实例化 Storyboard，您可以这样做：
let storyboard: UIStoryboard? = self.storyboard // call this inside a VC that is on the Storyboard
Note that in the last case, self.storyboardwill return an optionalStoryboard (Storyboard?), so if you'd like to use it unwrap it like so:
请注意，在最后一种情况下，self.storyboard将返回一个可选的Storyboard ( Storyboard?)，因此如果您想使用它，请像这样解开它：
if let storyboard = self.storyboard {
// access storyboard here
}
回答by Imran Khan
I have just copy pasted the code form above updated question so that everyone can see it as an answer.
我刚刚复制粘贴了更新问题上方的代码表单，以便每个人都可以将其视为答案。
UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];
