How to correctly use ExtractAction for inferred type on RunEvery

[Steven MacCoun]

Not sure if just need more documentation for ExtractAction, but I can't seem to get the right action type from my actionCreators for RunEvery.

const someAction = actionFactory<number>('SomeAction')
type SomeAction = ExtractAction<typeof someAction>


const setRetrieveState: RunEvery<any, SomeAction> = (state, action) => {
  const newValue = action.value + 1 // -- | ERROR! Object is of type unknown
  return from([{type: 'SettingSomeRetrieveState', value: newValue}])
}

Where typescript infers SomeAction as

type SomeAction = Action<unknown, Record<string, any>>

[Brandon Blaylock]

Ah, this is because actionFactory returns an ActionFunction, not an ActionCreator. The difference between the two is that ActionFunction only creates actions while ActionCreators have the match function. Here is one way to resolve the type problem:

import * as S from './src/Store';
import * as A from './src/Actions';
import { from } from 'rxjs';

const someActionGroup = A.actionCreatorFactory("GROUP");
const someAction = someActionGroup.simple<number>("SOME_ACTION")
type SomeAction = A.ExtractAction<typeof someAction>


const setRetrieveState: S.RunEvery<any, SomeAction> = (state, action) => {
  const newValue = action.value + 1;
  return from([{type: 'SettingSomeRetrieveState', value: newValue}])
}

You could also use the filterEvery combinator from Store.ts:

import * as S from './src/Store';
import * as A from './src/Actions';
import { from } from 'rxjs';

const someActionGroup = A.actionCreatorFactory("GROUP");
const someAction = someActionGroup.simple<number>("SOME_ACTION")

const setRetrieveState = S.filterEvery(someAction, (_, action) => {
  const newValue = action.value + 1;
  return from([{type: 'SettingSomeRetrieveState', value: newValue}])
});

Hope this helps!

[Steven MacCoun]

That is perfect, thank you!

[Brandon Blaylock]

Closing as the answer was accepted.