TypeScript 包装器函数推断类型

Question

我对与数据库操作相关的函数使用了一个小实用程序/包装函数。它应该采用类型为Promise<PromiseReturnType<GENERIC>> 的特定函数fn，然后在执行之前做一些事情（这里：检查连接）：

// File `A.ts`
type PromiseReturnType<T = undefined> = {success: boolean, response: T } 

const DatabaseOperation = <Fn extends (...args: any) => Promise<PromiseReturnType<Data>>, Data>(fn: (...args: any) => Promise<PromiseReturnType<Data>>) => {

    return async function (...args: Parameters<Fn>): Promise<PromiseReturnType<Data>> {
        if (!isConnected) return Promise.reject({ success: false, response: new Error("Not connected to the Database!") })

        return await fn(...args)
    }

}

然后我像这样使用它：

// File `B.ts`
const dbOperation = async (myArg: number): Promise<PromiseReturnType<string> => {

    return Promise.resolve({ success: true, response: `hello world ${myArg}` })

}


export default DatabaseOperation<typeof dbOperation, string>(dbOperation)

除了无法推断返回类型外，它工作正常。除了Promise<PromiseReturnType<string>>，它将是Promise<any>。

使用新包装的函数将如下所示：

import dbOperation from "B.ts"

(function () {
   await dbOperation() // Type: dbOperation(myArg: number): Promise<any>
})()

关于如何解决这个问题或总体上更好地实施这个问题有什么建议吗？

感谢您的帮助，谢谢。

Answer 1

你快到了：

declare var isConnected: boolean;

type PromiseReturnType<T = undefined> = { success: boolean, response: T }

const DatabaseOperation = <Data, Fn extends (...args: any[]) => Promise<PromiseReturnType<Data>>,>(fn: (...args: any[]) => Promise<PromiseReturnType<Data>>) => {

    return async function (...args: Parameters<Fn>): Promise<PromiseReturnType<Data>> {
        if (!isConnected) return Promise.reject({ success: false, response: new Error("Not connected to the Database!") })

        return await fn(...args)
    }

}

const dbOperation = (myArg: number): Promise<PromiseReturnType<string>> =>
    Promise.resolve({ success: true, response: `hello world ${myArg}` })



const fn = DatabaseOperation(dbOperation)

const result = fn(42) // Promise<PromiseReturnType<string>>

Playground

您应该将...args 设为数组any[] 而不仅仅是any