TypeScript type definition for promise.reject

问题

The following code is correct in terms of the type that is returned, because then always return the promise array.

Promise.resolve(['one', 'two'])
.then( arr =>
{
  if( arr.indexOf('three') === -1 )
    return Promise.reject( new Error('Where is three?') );

  return Promise.resolve(arr);
})
.catch( err =>
{
  console.log(err); // Error: where is three?
})

TypeScript throw error:

The type argument for type parameter 'TResult' cannot be inferred from the usage. Consider specifying the type arguments explicitly. Type argument candidate 'void' is not a valid type argument because it is not a supertype of candidate 'string[]'.

But in reality, then never will return void.

I can explicitly specify type .then<Promise<any>>, but it's more like a workaround, not the right solution.

How to write this right?

回答1:

You should not return Promise.resolve and Promise.reject inside a promise chain. The resolve should be driven by simple return and reject should be driven by an explicit throw new Error.

Promise.resolve(['one', 'two'])
.then( arr =>
{
  if( arr.indexOf('three') === -1 )
    throw new Error('Where is three?');

  return arr;
})
.catch( err =>
{
  console.log(err); // Error: where is three?
})

More on promise chaining https://basarat.gitbooks.io/typescript/content/docs/promise.html

回答2:

Typescript is complaining about the difference in return type between your Promise.reject return value (Promise<void>) and your Promise.resolve value (Promise<string[]>).

Casting your then call as .then<Promise<void | string[]>> will let the compiler know of the union return type.

as @basarat notes, you should just throw an error instead of using Promise.reject (which will be passed to whatever catch handler is provided).

来源：https://stackoverflow.com/questions/39381940/typescript-type-definition-for-promise-reject

标签

typescript

es6-promise