Question

TL; DR：我的函数接受构造函数或构造对象，对其进行变异，然后返回它。如何正确键入返回值，以便类保持类，对象保留对象，并且两者都有新属性？

详情

我关闭：当混合到对象中时，原始方法和混合方法都可用，并且当混合到构造函数中时，只有混合方法可用。

我目前的解决方案有两个问题：

当混合到构造函数中时，生成的实例的类型是MixinType & typof OriginalClass而不是MixinType & OriginalClass（OriginalClass上的静态方法可用作实例方法，而实例方法不是{＆1;可用
当混合到构造函数中时，构造函数在其类型签名中没有参数，它具有MixinTypeClass的构造函数签名（这并不令人惊讶，这就是我写它的方式 - 这就是为什么我要问这个问题，我正在寻找另一种方法来做这件事。

我最好的尝试

class OriginalClass {
    instanceMethod(): void { }
}

interface MixinType {
    mixedIn(): void;
}

interface MixinTypeClass<T> {
    new (): MixinType & T;
}

function Mixin<T extends Function>(classOrInstance: T): MixinTypeClass<T>;
function Mixin<T extends Object>(classOrInstance: T): T & MixinType;
function Mixin<T extends any>(classOrInstance: T): any {
    if (typeof classOrInstance === 'function') {
        (classOrInstance.prototype as any).mixedIn = () => { }
        return classOrInstance as any as MixinTypeClass<T>;
    } else {
        (classOrInstance as any).mixedIn = () => { }
        return classOrInstance as typeof classOrInstance & MixinType;
    }
}

let NewClass = Mixin(OriginalClass);
let instance1 = new NewClass();
instance1.mixedIn();
// Property 'instanceMethod' does not exist on the type 'MixinType & typeof OriginalClass'
instance1.instanceMethod();

let instance2 = Mixin(new OriginalClass());
instance2.mixedIn();
instance2.instanceMethod();

那么，如何更改Mixin以便instance1.instanceMethod()没有编译器错误？

Answer 1

将var org = [ {name:"one",age:1} ,{name:"two",age:2} ] ; var newArray = org .map( (x,index)=> index === 1 ?Object.assign( {} ,x ,{name:"new name"} ) :x ) ;缩短为简单的classOrInstance，你可以这样做：

arg

这提供了完整的智能感知和类型安全：

function Mixin<T extends Object>(arg: Function): MixinTypeClass<T>;
function Mixin<T extends Object>(arg: T): T & MixinType;
function Mixin<T extends Object>(arg: Function | T): any {
    if (arg instanceof Function) {
        (arg.prototype as any).mixedIn = () => { }
        return arg as MixinTypeClass<T>;
    } else {
        (arg as any).mixedIn = () => { }
        return arg as T & MixinType;
    }
}

值得注意的是，let NewClass: MixinTypeClass<OriginalClass> = Mixin<OriginalClass>(OriginalClass); let instance1: OriginalClass & MixinType = new NewClass(); instance1.mixedIn(); instance1.instanceMethod(); let instance2: OriginalClass & MixinType = Mixin<OriginalClass>(new OriginalClass()); instance2.mixedIn(); instance2.instanceMethod();和MixinTypeClass<OriginalClass>并不相同。

键入可应用于类和实例的Typescript mixin函数

1 个答案: