object 中的类型从异步 function 返回 Typescript

Question

Given a function that returns an object promise with a boolean & string:

const test = () => {
   return new Promise(async (resolve) => {
       resolve({ result: false, error: 'This is an error' });
   })
}

I try to destruct these couple of values into constants:

const { result, error } = await test();

However, I always get these Typescript errors:

Property 'result' does not exist on type 'unknown'.ts(2339)
Property 'error' does not exist on type 'unknown'.ts(2339)

I've tried all kinds of combinations, but the only one working is adding the 'any' type, which I believe we should always avoid.

Any idea to define the right types without getting error?

Answer 1

Here you should add the type as the generic parameter to the promise - ie new Promise<MyType>(...) Example:

type MyPromiseResult = { result: boolean, error: string };

const test = () => {
   return new Promise<MyPromiseResult>((resolve) => {
       resolve({ result: false, error: 'This is an error' });
   })
}

async function doThing() {
  const { result, error } = await test();
  type ResultType = typeof result; // boolean
  type ErrorType = typeof error; // string
}

Answer 2

Give your response a type

return new Promise<{ result: boolean, error: string }>((resolve) => {})