如何在swift 3中从字符串创建类的实例

Question

我尝试使用字符串以多种方式创建类的实例,但没有一个在Swift 3中工作.

下面是我试过的Swift 3之前的解决方案,它们无效

- 使课堂成为一个客观的课程

@objc(customClass)
class customClass {
    ...
}

//Error here: cannot convert value of type 'AnyClass?' to expected argument type 'customClass'
let c: customClass = NSClassFromString("customClass")

- 使用NSString值指定类(使用和不使用@objc属性)

@objc(customClass)
class customClass {
    ...
}

//Error here: cannot convert value of type 'String' to expected argument type 'AnyClass' (aka 'AnyObject.Type')
var className = NSStringFromClass("customClass")

let c: customClass = NSClassFromString(className)

我做的不对,但没有在网上找到任何解决方案.

如何在Swift 3中使用字符串创建类的实例？

Answer 1

你可以试试这个:

func stringClassFromString(_ className: String) -> AnyClass! {

    /// get namespace
    let namespace = Bundle.main.infoDictionary!["CFBundleExecutable"] as! String;

    /// get 'anyClass' with classname and namespace 
    let cls: AnyClass = NSClassFromString("\(namespace).\(className)")!;

    // return AnyClass!
    return cls;
}

像这样使用func:

class customClass: UITableView {

    }   
let myclass = stringClassFromString("customClass") as! UITableView.Type
let instance = myclass.init()

Answer 2

如果您的项目包含空格，则应将空格替换为“_”

像这样的东西

let namespace = (Bundle.main.infoDictionary!["CFBundleExecutable"] as! String).replacingOccurrences(of: " ", with: "_")
let cls = NSClassFromString("\(namespace).\(className)")! as! AnyClass

Answer 3

苹果提供了一种无需使用即可实现此目的的方法NSClassFromString。

Bundle.main.classNamed("MyClassName")

https://developer.apple.com/documentation/foundation/bundle/1407299-classnamed