正确使用 reduce-reducers

Question

我不明白reduce-reducers 是什么意思。如果我有 2 个包含相同操作的 reducer 函数，是否应该使用它？

function reducerA(state, action){
   switch(action.type):
       ...
       case 'SAME_ACTION': {...state, field: state.field+1}
}

function reducerB(state, action){
   switch(action.type):
       ...
       case 'SAME_ACTION': {...state, field: state.field*2}
}

因此，如果我在 reducerA 和 reducerB 上调用 reduceReducer 并为 {field: 0} 调用操作“SAME_ACTION”，那么我将有下一个状态 {field: 2}？

在我看来，它有点连接化简器（意味着将它们合并到一个键下）。

我是对的还是reduceReducer 有不同的目的？

Answer 1

区别在于：

• combineReducers 创建嵌套状态
• reduceReducers 创建 flat 状态

考虑以下减速器。没有动作类型可以让事情变得更简单：

// this reducer adds a payload to state.sum 
// and tracks total number of operations
function reducerAdd(state, payload) {
  if (!state) state = { sum: 0, totalOperations: 0 }
  if (!payload) return state

  return {
    ...state,
    sum: state.sum + payload,
    totalOperations: state.totalOperations + 1
  }
}

// this reducer multiplies state.product by payload
// and tracks total number of operations
function reducerMult(state, payload) {
  if (!state) state = { product: 1, totalOperations: 0 }
  if (!payload) return state

  // `product` might be undefined because of 
  // small caveat in `reduceReducers`, see below
  const prev = state.product || 1

  return {
    ...state,
    product: prev * payload,
    totalOperations: state.totalOperations + 1
  }
}

组合减速器

每个reducer 都有一个独立的状态（另见http://redux.js.org/docs/api/combineReducers.html）：

const rootReducer = combineReducers({
  add: reducerAdd,
  mult: reducerMult
})

const initialState = rootReducer(undefined)
/*
 * {
 *   add:  { sum: 0, totalOperations: 0 },
 *   mult: { product: 1, totalOperations: 0 },
 * }
 */


const first = rootReducer(initialState, 4)
/*
 * {
 *   add:  { sum: 4, totalOperations: 1 },
 *   mult: { product: 4, totalOperations: 1 },
 * }
 */    
// This isn't interesting, let's look at second call...

const second = rootReducer(first, 4)
/*
 * {
 *   add:  { sum: 8, totalOperations: 2 },
 *   mult: { product: 16, totalOperations: 2 },
 * }
 */
// Now it's obvious, that both reducers get their own 
// piece of state to work with

reduceReducers

所有reducer共享相同的状态

const addAndMult = reduceReducers(reducerAdd, reducerMult) 

const initial = addAndMult(undefined)
/* 
 * {
 *   sum: 0,
 *   totalOperations: 0
 * }
 *
 * First, reducerAdd is called, which gives us initial state { sum: 0 }
 * Second, reducerMult is called, which doesn't have payload, so it 
 * just returns state unchanged. 
 * That's why there isn't any `product` prop.
 */ 

const next = addAndMult(initial, 4)
/* 
 * {
 *   sum: 4,
 *   product: 4,
 *   totalOperations: 2
 * }
 *
 * First, reducerAdd is called, which changes `sum` = 0 + 4 = 4
 * Second, reducerMult is called, which changes `product` = 1 * 4 = 4
 * Both reducers modify `totalOperations`
 */


const final = addAndMult(next, 4)
/* 
 * {
 *   sum: 8,
 *   product: 16,
 *   totalOperations: 4
 * }
 */

用例

• combineReducers - 每个 reducer 管理自己的状态切片（例如 state.todos 和 state.logging）。这在创建 root reducer 时很有用。
• reduceReducers - 每个 reducer 管理相同的状态。这在链接多个应该在相同状态下运行的减速器时很有用（例如，当组合使用 redux-actions 中的 handleAction 创建的多个减速器时，可能会发生这种情况）

从最终的状态形状来看，区别很明显。

Answer 2

我也不明白 reduce-reducers 试图解决的问题。 @Tomáš 描述的用例可以通过一个简单的 Reducer 来实现。毕竟，Reducer 只是一个接受 app-state 和 action 的函数，并返回一个包含新 app-state 的对象。例如，您可以执行以下操作，而不是使用 redux 提供的 combineReducers：

import combinationReducer from "./combinationReducer";
import endOfPlayReducer from "./endOfPlayReducer";
import feedbackReducer from "./feedbackReducer";

function combineReducers(appState, action) {
  return {
    combination: combinationReducer(appState, action),
    feedbacks: feedbackReducer(appState, action),
    endOfPlay: endOfPlayReducer(appState, action)
  };
}

当然，在这里，您的 reducer 接受整个应用程序状态并仅返回它们负责的切片。同样，它只是一个功能，您可以随意自定义它。你可以阅读更多关于它here