在Typescript中创建通用方法装饰器

Question

在这段代码中,我试图让用户为通用装饰器函数fetchJson指定一个类型.

它会像这样工作:

1. 用以下方法装饰方法: @fetchJson<User>
2. 然后我们用一个自动调用的函数替换该函数.then(res => res.json()),并返回一个包含在Promise中的类型值.

我遇到的问题是我不知道如何将返回分配给descriptor.value用户指定的T.是否有更好的方法来执行此操作？我觉得我完全错过了一些东西.

interface PromiseDescriptorValue<T>{
  (...args: any[]): Promise<T>;
}

const fetchJson = <T>(
  target: Object,
  propertyKey: string,
  descriptor: TypedPropertyDescriptor<PromiseDescriptorValue<Response>> // Response is a whatwg-fetch response -- https://github.com/DefinitelyTyped/DefinitelyTyped/blob/master/whatwg-fetch/whatwg-fetch.d.ts#L58
): TypedPropertyDescriptor<PromiseDescriptorValue<T>> => {
  const oldMethod = descriptor.value;

  descriptor.value = function(...args: any[]) {
    return oldMethod.apply(this, args).then((res: Response) => res.json());
  };

  return descriptor;
};


// TS2322: Type 'TypedPropertyDescriptor<PromiseDescriptorValue<Response>>'
// is not assignable to type
// 'TypedPropertyDescriptor<PromiseDescriptorValue<T>>'. Type
// 'PromiseDescriptorValue<Response>' is not assignable to type
// 'PromiseDescriptorValue<T>'. Type 'Response' is not assignable to type 'T'.

Answer 1

也许是这样的:

interface PromiseDescriptorValue<T>{
  (...args: any[]): Promise<T>;
}

export function fetchJson<T>(): any
{
    return (
        target: Object,
        propertyKey: string,
        descriptor: any): TypedPropertyDescriptor<PromiseDescriptorValue<T>> => 
        {
        const oldMethod = descriptor.value;

        descriptor.value = function(...args: any[]) 
        {
            return oldMethod.apply(this, args).then((res: Response) => res.json());
        };

        return descriptor;
        };
}

class C 
{
    @fetchJson<User>()
    foo(args) 
    {
        //....
    }
}