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？

Question

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？我查看了应用程序代表的参考,但到目前为止我还没有找到一个.

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

这很简单.当我遇到类似的问题时,我编写了一个可以从主bundle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
    }
}

这是一个简单的例子:

//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的答案是正确的,也有一些情况下你根本没有对a的引用UIViewController,所以你可以从你rootViewController的UIWindow对象中获取它AppDelegate.

// 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

当然,您也可以UIStoryboard使用(如@Mario建议的那样)创建一个:

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

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:捆绑:)

为指定的Storyboard资源文件创建并返回storyboard对象.

init(name: String, bundle storyboardBundleOrNil: Bundle?)

参数

• name: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.
• storyboardBundleOrNil:包含故事板文件及其相关资源的捆绑包.如果指定nil,则此方法将在当前应用程序的主包中查找.

来源:Apple文档