@objc(SwiftClassName)
。@objc(SubClass)
class SubClass: SuperClass {...}
NSClassFromString()
函数需要由 @objc
属性指定的名称。 - eonil@objc(SubClass)
有效,而@objc class SubClass
无效的问题? - pyanfield@objc class SubClass
形式中,名称被暗示为与 SubClass 名称相同。而在 @objc(SubClass) class SubClass
形式中,则直接指定。我想编译器无法自行解决第一种形式的问题。 - voiger以下是我按照类名初始化派生的UIViewController的方式
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass()
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass.init()
这里有一个更好的解决方案:https://dev59.com/AWAg5IYBdhLWcg3wDHU3#32265287
请注意,Swift类现在是有命名空间的,所以不再使用“MyViewController”,而是使用“AppName.MyViewController”
XCode6-beta 6/7已经被弃用
解决方案是在XCode6-beta 3中开发的
由于Edwin Vermeer的答案,我能够构建一些东西来将Swift类实例化为Obj-C类,方法如下:
// 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
你也可以使用纯 Obj-C 完成它:
- (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);
}
我希望这能帮助到某个人!
更新:从beta 6开始,NSStringFromClass将返回由您的包名称和类名称用点分隔的字符串。因此,它将类似于MyApp.MyClass。
Swift类将具有以下部分组成的构造内部名称:
因此,您的类名将类似于_TtC5MyApp7MyClass。
您可以通过执行以下操作将其作为字符串获取:
var classString = NSStringFromClass(self.dynamicType)
更新 在Swift 3中已经更改为:
var classString = NSStringFromClass(type(of: self))
使用该字符串,您可以通过执行以下操作创建Swift类的实例:
var anyobjectype : AnyObject.Type = NSClassFromString(classString)
var nsobjectype : NSObject.Type = anyobjectype as NSObject.Type
var rec: AnyObject = nsobjectype()
差不多一样
func NSClassFromString(_ aClassName: String!) -> AnyClass!
请查看此文档:
NSClass
类,而不适用于Swift类。 NSClassFromString(“String”)
返回nil
,但NSClassFromString(“NSString”)
不会返回nil
。 - Cezary Wojcikvar myVar:NSClassFromString("myClassName")
- gabuhString
不是一个类,它是一个结构体。 - newacctNSClassFromString
也会对所有的Swift类返回nil
。 - Cezary Wojcikvar clazz: NSObject.Type = TestObject.self
var instance : NSObject = clazz()
if let testObject = instance as? TestObject {
println("yes!")
}
AnyClass
)。我认为他们不希望你这样做,因为这基本上会破坏类型系统。针对Swift2,我创建了一个非常简单的扩展程序,以更快地实现此操作 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()
}
}
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)
}
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)
}
_stdlib_getTypeName
获取变量的编码类型名称。将其粘贴到空白playground中: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))")
TypeName0 = NSString
TypeName1 = _TtC13__lldb_expr_014PureSwiftClass
TypeName2 = _TtSi
TypeName3 = _TtSS
Ewan Swick的博客文章帮助解密这些字符串:http://www.eswick.com/2014/06/inside-swift/
例如:_TtSi
代表Swift内部的Int
类型。
xcode 7 beta 5:
class MyClass {
required init() { print("Hi!") }
}
if let classObject = NSClassFromString("YOURAPPNAME.MyClass") as? MyClass.Type {
let object = classObject.init()
}