如何从 typesafe redux-observable epic 中收听“@@router/LOCATION_CHANGE”动作答案

【问题标题】：How do I listen to '@@router/LOCATION_CHANGE' action from typesafe redux-observable epic如何从 typesafe redux-observable epic 中收听“@@router/LOCATION_CHANGE”动作
【发布时间】：2021-10-08 11:22:37
【问题描述】：

我正在尝试收听来自 typesafe redux-observable epic 的 '@@router/LOCATION_CHANGE' 动作，但我只是不明白如何。

我需要过滤的具体操作是什么？我尝试'filter(onLocationChanged)' 失败。
什么是正确的'IN' 类型？使用 'LocationChangeAction' 和 'filter(onLocationChanged)' 我得到类型错误：

没有重载匹配这个调用。

最后，如何通过'@@router/LOCATION_CHANGE' 操作更改位置？（之前和之后）

import { Epic } from "redux-observable";
import { of } from "rxjs";
import { filter, withLatestFrom, concatMap } from "rxjs/operators";
import { AppState } from "../../reducer";
import { CallHistoryMethodAction, LOCATION_CHANGE, push, onLocationChanged, RouterActionType, LocationChangeAction } from 'connected-react-router';


type IN = LocationChangeAction;
type OUT = ReturnType<typeof somthing>;

export const LocationListenEpic: Epic<IN | OUT, OUT, AppState> = (
    action$,
    state$
) => 
    action$.pipe(
        filter(onLocationChanged),
        withLatestFrom(state$),
        concatMap(([{ payload }, state]) => {

            console.log(currentLocation);
            return []
}

【问题讨论】：

标签： typescript react-router redux-observable

【解决方案1】：

在深入了解 redux-observable epics 的工作原理后：


import { getLocation } from 'connected-react-router'

type IN = LocationChangeAction;
type OUT = ReturnType<typeof somthing>;

export const LocationListenEpic: Epic<IN | OUT, OUT, AppState> = (
    action$,
    state$
) => 
    action$.pipe(
        filter(action => action.type == LOCATION_CHANGE),
        withLatestFrom(state$),
        concatMap(([{ payload }, state]) => {

            console.log(getLocation(store.getState()));
            return []
}

【讨论】：