ios Swift 语言 NSClassFromString

Swift language NSClassFromString

提问by Selvin

How to achieve reflectionin Swift Language?

如何在 Swift 语言中实现反射？

How can I instantiate a class

如何实例化一个类

[[NSClassFromString(@"Foo") alloc] init];

采纳答案by Kevin Delord

Less hacky solution here: https://stackoverflow.com/a/32265287/308315

这里不那么hacky的解决方案：https: //stackoverflow.com/a/32265287/308315

Note that Swift classes are namespaced now so instead of "MyViewController" it'd be "AppName.MyViewController"

请注意，Swift 类现在已命名空间，因此不是“MyViewController”，而是“AppName.MyViewController”

---

---

Deprecated since XCode6-beta 6/7

Solution developed using XCode6-beta 3

自 XCode6-beta 6/7 起已弃用

使用 XCode6-beta 3 开发的解决方案

Thanks to the answer of Edwin Vermeer I was able to build something to instantiate Swift classes into an Obj-C class by doing this:

感谢 Edwin Vermeer 的回答，我能够通过这样做来构建一些东西来将 Swift 类实例化为 Obj-C 类：

// swift file
// extend the NSObject class
extension NSObject {
    // create a static method to get a swift class for a string name
    class func swiftClassFromString(className: String) -> AnyClass! {
        // get the project name
        if  var appName: String? = NSBundle.mainBundle().objectForInfoDictionaryKey("CFBundleName") as String? {
            // generate the full name of your class (take a look into your "YourProject-swift.h" file)
            let classStringName = "_TtC\(appName!.utf16count)\(appName)\(countElements(className))\(className)"
            // return the class!
            return NSClassFromString(classStringName)
        }
        return nil;
    }
}

// obj-c file
#import "YourProject-Swift.h"

- (void)aMethod {
    Class class = NSClassFromString(key);
    if (!class)
        class = [NSObject swiftClassFromString:(key)];
    // do something with the class
}

EDIT

编辑

You can also do it in pure obj-c:

您也可以在纯 obj-c 中执行此操作：

- (Class)swiftClassFromString:(NSString *)className {
    NSString *appName = [[NSBundle mainBundle] objectForInfoDictionaryKey:@"CFBundleName"];
    NSString *classStringName = [NSString stringWithFormat:@"_TtC%d%@%d%@", appName.length, appName, className.length, className];
    return NSClassFromString(classStringName);
}

I hope this will help somebody !

我希望这会帮助某人！

回答by Rigel Chen

This is the way I init derived UIViewController by class name

这是我通过类名初始化派生 UIViewController 的方式

var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass()

More information is here

更多信息在这里

In iOS 9

在 iOS 9 中

var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass.init()

回答by klaus

You must put @objc(SwiftClassName)above your swift class.
Like:

你必须把@objc(SwiftClassName)你的 swift 类放在上面。
喜欢：

@objc(SubClass)
class SubClass: SuperClass {...}

回答by Edwin Vermeer

UPDATE: Starting with beta 6 NSStringFromClass will return your bundle name plus class name separated by a dot. So it will be something like MyApp.MyClass

更新：从 beta 6 NSStringFromClass 开始将返回您的包名称加上用点分隔的类名称。所以它将类似于 MyApp.MyClass

Swift classes will have a constructed internal name that is build up of the following parts:

Swift 类将有一个由以下部分构成的构造内部名称：

• It will start with _TtC,
• followed by a number that is the length of your application name,
• followed by your application name,
• folowed by a number that is the length of your class name,
• followed by your class name.

• 它将以 _TtC 开头，
• 后跟一个数字，表示您的应用程序名称的长度，
• 后跟您的应用程序名称，
• 后跟一个数字，它是您的班级名称的长度，
• 后跟您的班级名称。

So your class name will be something like _TtC5MyApp7MyClass

所以你的类名将类似于 _TtC5MyApp7MyClass

You can get this name as a string by executing:

您可以通过执行以下命令以字符串形式获取此名称：

var classString = NSStringFromClass(self.dynamicType)

UpdateIn Swift 3 this has changed to:

更新在 Swift 3 中，这已更改为：

var classString = NSStringFromClass(type(of: self))

Using that string, you can create an instance of your Swift class by executing:

使用该字符串，您可以通过执行以下命令来创建 Swift 类的实例：

var anyobjectype : AnyObject.Type = NSClassFromString(classString)
var nsobjectype : NSObject.Type = anyobjectype as NSObject.Type
var rec: AnyObject = nsobjectype()

回答by gabuh

It's almost the same

几乎一样

func NSClassFromString(_ aClassName: String!) -> AnyClass!

Check this doc:

检查此文档：

https://developer.apple.com/library/prerelease/ios/documentation/Cocoa/Reference/Foundation/Miscellaneous/Foundation_Functions/#//apple_ref/c/func/NSClassFromString

https://developer.apple.com/library/prerelease/ios/documentation/Cocoa/Reference/Foundation/Miscellaneous/Foundation_Functions/#//apple_ref/c/func/NSClassFromString

回答by Sulthan

I was able to instantiate an object dynamically

我能够动态实例化一个对象

var clazz: NSObject.Type = TestObject.self
var instance : NSObject = clazz()

if let testObject = instance as? TestObject {
    println("yes!")
}

I haven't found a way to create AnyClassfrom a String(without using Obj-C). I think they don't want you to do that because it basically breaks the type system.

我还没有找到AnyClass从String（不使用 Obj-C）创建的方法。我认为他们不希望你这样做，因为它基本上打破了类型系统。

回答by Damien Romito

For swift2, I created a very simple extension to do this more quicklyhttps://github.com/damienromito/NSObject-FromClassName

对于 swift2，我创建了一个非常简单的扩展来更快地执行此操作https://github.com/damienromito/NSObject-FromClassName

extension NSObject {
    class func fromClassName(className : String) -> NSObject {
        let className = NSBundle.mainBundle().infoDictionary!["CFBundleName"] as! String + "." + className
        let aClass = NSClassFromString(className) as! UIViewController.Type
        return aClass.init()
    }
}

---

---

In my case, i do this to load the ViewController I want:

就我而言，我这样做是为了加载我想要的 ViewController：

override func viewDidLoad() {
    super.viewDidLoad()
    let controllers = ["SettingsViewController", "ProfileViewController", "PlayerViewController"]
    self.presentController(controllers.firstObject as! String)

}

func presentController(controllerName : String){
    let nav = UINavigationController(rootViewController: NSObject.fromClassName(controllerName) as! UIViewController )
    nav.navigationBar.translucent = false
    self.navigationController?.presentViewController(nav, animated: true, completion: nil)
}

回答by Go Let

In Swift 2.0 (tested in the Xcode 7.01) _20150930

在 Swift 2.0（在 Xcode 7.01 中测试）_20150930

let vcName =  "HomeTableViewController"
let ns = NSBundle.mainBundle().infoDictionary!["CFBundleExecutable"] as! String

// Convert string to class
let anyobjecType: AnyObject.Type = NSClassFromString(ns + "." + vcName)!
if anyobjecType is UIViewController.Type {
// vc is instance
    let vc = (anyobjecType as! UIViewController.Type).init()
    print(vc)
}

回答by Klaas

This will get you the name of the class that you want to instantiate. Then you can use Edwins answerto instantiate a new object of your class.

这将为您提供要实例化的类的名称。然后你可以使用Edwins answer来实例化你的类的一个新对象。

As of beta 6 _stdlib_getTypeNamegets the mangled type name of a variable. Paste this into an empty playground:

从 beta 6 开始，_stdlib_getTypeName获取变量的重整类型名称。将其粘贴到一个空的操场上：

import Foundation

class PureSwiftClass {
}

var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"

println( "TypeName0 = \(_stdlib_getTypeName(myvar0))")
println( "TypeName1 = \(_stdlib_getTypeName(myvar1))")
println( "TypeName2 = \(_stdlib_getTypeName(myvar2))")
println( "TypeName3 = \(_stdlib_getTypeName(myvar3))")

The output is:

输出是：

TypeName0 = NSString
TypeName1 = _TtC13__lldb_expr_014PureSwiftClass
TypeName2 = _TtSi
TypeName3 = _TtSS

Ewan Swick's blog entry helps to decipher these strings: http://www.eswick.com/2014/06/inside-swift/

Ewan Swick 的博客条目有助于破译这些字符串：http: //www.eswick.com/2014/06/inside-swift/

e.g. _TtSistands for Swift's internal Inttype.

eg_TtSi代表 Swift 的内部Int类型。

回答by December

xcode 7 beta 5:

Xcode 7 测试版 5：

class MyClass {
    required init() { print("Hi!") }
}
if let classObject = NSClassFromString("YOURAPPNAME.MyClass") as? MyClass.Type {
    let object = classObject.init()
}