Swift语言NSClassFromString

Question

如何在Swift语言中实现反思？

我如何实例化一个类

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

Answer 1

这是我通过类名初始化UIViewController的方式

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

更多信息在这里

在iOS 9中

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

Answer 2

你必须把@objc(SwiftClassName)你的快速课程放在上面.
喜欢:

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

Answer 3

这里不太讨厌的解决方案:https://stackoverflow.com/a/32265287/308315

请注意,Swift类现在是命名空间,因此它不是"MyViewController"而是"AppName.MyViewController"

自XCode6-beta 6/7后不再使用

使用XCode6-beta 3开发的解决方案

感谢Edwin Vermeer的回答,我能够构建一些东西,通过这样做将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
}

编辑

你也可以用纯粹的obj-c来做:

- (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);
}

我希望这会对某人有所帮助!

Answer 4

更新:从beta 6开始NSStringFromClass将返回您的包名称加上以点分隔的类名.所以它会像MyApp.MyClass

Swift类将具有由以下部分构成的构造内部名称:

• 它将从_TtC开始,
• 后跟一个数字,即应用程序名称的长度,
• 然后是您的应用程序名称,
• 下面是一个数字,即你的班级名称的长度,
• 然后是你的班级名字.

所以你的类名将是_TtC5MyApp7MyClass

您可以通过执行以下命令将此名称作为字符串:

var classString = NSStringFromClass(self.dynamicType)

更新在Swift 3中,这已更改为:

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

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()

Answer 5

它几乎是一样的

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

检查此文档:

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

Answer 6

我能够动态地实例化一个对象

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

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

我还没有找到一种方式来创建AnyClass从一个String(不使用的OBJ-C).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.

Answer 7

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()
    }
}

在我的情况下,我这样做加载我想要的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)
}

Answer 8

这将为您提供要实例化的类的名称.然后,您可以使用Edwins的答案来实例化您的类的新对象.

从beta 6开始,_stdlib_getTypeName获取变量的错位类型名称.将其粘贴到空的操场上:

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))")

输出是:

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

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

例如_TtSi代表Swift的内部Int类型.

Answer 9

在Swift 2.0中(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)
}

Answer 10

xcode 7 beta 5:

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