斯威夫特3.0: 你可以创建一个像这样的扩展..它返回的是类名，而不是项目名

extension NSObject {
    var className: String {
        return NSStringFromClass(self as! AnyClass).components(separatedBy: ".").last ?? ""
    }

    public class var className: String {
        return NSStringFromClass(self).components(separatedBy: ".").last ?? ""
    }
}

你可以像这样使用Swift标准库函数_stdlib_getDemangledTypeName:

let name = _stdlib_getDemangledTypeName(myViewController)

这类例子用于类var.不要包含bundle的名称。

extension NSObject {
    class var className: String {
        return "\(self)"
    }
}

试着反映了()。类self或实例dynamicType的摘要。在获取dynamicType之前解开可选项，否则dynamicType是可选包装器。

class SampleClass { class InnerClass{} }
let sampleClassName = reflect(SampleClass.self).summary;
let instance = SampleClass();
let instanceClassName = reflect(instance.dynamicType).summary;
let innerInstance = SampleClass.InnerClass();
let InnerInstanceClassName = reflect(innerInstance.dynamicType).summary.pathExtension;
let tupleArray = [(Int,[String:Int])]();
let tupleArrayTypeName = reflect(tupleArray.dynamicType).summary;

摘要是描述了泛型类型的类路径。要从摘要中获得简单的类名，请尝试此方法。

func simpleClassName( complexClassName:String ) -> String {
    var result = complexClassName;
    var range = result.rangeOfString( "<" );
    if ( nil != range ) { result = result.substringToIndex( range!.startIndex ); }
    range = result.rangeOfString( "." );
    if ( nil != range ) { result = result.pathExtension; }
    return result;
}

要获得类名为String，请按以下方式声明类

@objc(YourClassName) class YourClassName{}

并使用以下语法获取类名

NSStringFromClass(YourClassName)

要在Swift 4中获得类型名为字符串(我没有检查早期版本)，只需使用字符串插值:

"\(type(of: myViewController))"

你可以在类型本身上使用.self，在实例上使用type(of:_)函数:

// Both constants will have "UIViewController" as their value
let stringFromType = "\(UIViewController.self)"
let stringFromInstance = "\(type(of: UIViewController()))"