Question

我如何以编程方式在swift中引用我的应用程序中的主要故事板？我查看了应用代表作为参考，但到目前为止我还没找到。

Answer 1

哦，哎呀，我找到了答案......

在另一个视图中，控制器连接到故事板，您只需使用：

self.storyboard?

Answer 2

或者任何对象都可以通过引用其名称和包来获取故事板：

let storyboard = UIStoryboard(name: "storyboardNameHere", bundle: nil) //if bundle is nil the main bundle will be used

Answer 3

这很容易。当我遇到类似的问题时，我写了一个可以从主包中获取任何资源的类。

//Generate name of the main storyboard file, by default: "Main"
var kMainStoryboardName: String {
    let info = NSBundle.mainBundle().infoDictionary!

    if let value = info["TPMainStoryboardName"] as? String
    {
        return value
    }else{
        return "Main"
    }
}

public class TPBundleResources
{
    class func nib(name: String) -> UINib?
    {
        let nib = UINib(nibName: name, bundle: NSBundle.mainBundle());
        return nib
    }

    //Main storybord
    class func mainStoryboard() -> UIStoryboard
    {
        return storyboard(kMainStoryboardName)
    }

    class func storyboard(name: String) -> UIStoryboard
    {
        let storyboard = UIStoryboard(name: name, bundle: NSBundle.mainBundle())
        return storyboard
    }

    //Obtain file from main bundle by name and fileType
    class func fileFromBundle(fileName: String?, fileType: String?) -> NSURL?
    {
        var url: NSURL?

        if let path = NSBundle.mainBundle().pathForResource(fileName, ofType: fileType)
        {
            url = NSURL.fileURLWithPath(path)
        }

        return url
    }

    class func plistValue(key:String) -> AnyObject?
    {
        let info = NSBundle.mainBundle().infoDictionary!

        if let value: AnyObject = info[key]
        {
            return value
        }else{
            return nil
        }
    }
}


public extension TPBundleResources
{
    //Obtain view controller by name from main storyboard
    class func vcWithName(name: String) -> UIViewController?
    {
        let storyboard = mainStoryboard()
        let viewController: AnyObject! = storyboard.instantiateViewControllerWithIdentifier(name)
        return viewController as? UIViewController
    }

    class func vcWithName(storyboardName:String, name: String) -> UIViewController?
    {
        let sb = storyboard(storyboardName)
        let viewController: AnyObject! = sb.instantiateViewControllerWithIdentifier(name)
        return viewController as? UIViewController
    }

    //Obtain view controller by idx from nib
    class func viewFromNib(nibName: String, atIdx idx:Int) -> UIView?
    {
        let view =  NSBundle.mainBundle().loadNibNamed(nibName, owner: nil, options: nil)[idx] as! UIView
        return view
    }

    class func viewFromNib(nibName: String, owner: AnyObject, atIdx idx:Int) -> UIView?
    {
        let bundle = NSBundle(forClass: owner.dynamicType)
        let nib = UINib(nibName: nibName, bundle: bundle)
        let view = nib.instantiateWithOwner(owner, options: nil)[idx] as? UIView
        return view
    }

    class func viewFromNibV2(nibName: String, owner: AnyObject, atIdx idx:Int) -> UIView?
    {
        let view =  NSBundle.mainBundle().loadNibNamed(nibName, owner: owner, options: nil)[idx] as! UIView
        return view
    }
}

以下是一些简单的例子：

//Get a main storyboard
TPBundleResources.mainStoryboard()

//Get view controller form main storyboard
TPBundleResources.vcWithName("MyViewController")

//Get view from MyView.nib at index 0
TPBundleResources.viewFromNib("MyView", atIdx: 0)

//Get plist value by key
TPBundleResources.plistValue("key")

Answer 4

即使@ ManOfPanda的回答是正确的，但有些情况下您根本没有引用UIViewController，因此您可以从rootViewController获取来自UIWindow的{{1}}对象。

AppDelegate

当然，您也可以使用（如@Mario建议的那样）创建// First import your AppDelegate import AppDelegate // ... // Then get a reference of it. let appDelegate = UIApplication().shared.delegate as! AppDelegate // From there, get your UIStoryboard reference from the // rootViewController in your UIWindow let rootViewController = appDelegate.window?.rootViewController let storyboard = rootViewController?.storyboard：

UIStoryboard

但是，根据Apple文档，这将创建一个新的Storyboard实例（即使你已经有一个工作）。我总是喜欢使用现有的实例。

的init（名称：束：）

为指定的故事板资源文件创建并返回storyboard对象。

let storyboard = UIStoryboard(name: "storyboard", bundle:nil)

<强>参数


init(name: String, bundle storyboardBundleOrNil: Bundle?)：没有文件扩展名的storyboard资源文件的名称。如果此参数为nil，则此方法引发异常。

name：包含故事板文件及其相关资源的捆绑包。如果指定nil，则此方法将查找当前应用程序的主包。

Source: Apple documentation

Answer 5

UIStoryboard * mainStoryboard = [UIStoryboard storyboardWithName:@"Main" bundle:nil];

如何在swift中以编程方式引用我的应用程序中的主要故事板？

5 个答案:

的init（名称：束：）