如何在redux中的reducer方法中调用其他reducer？或者reducer中获取全局的state

在程序中我有两个reducer： clock与todo。
clock是个倒计时闹钟程序，
todo是任务管理。    
于是需要在执行clock的reducer修改todo中的状态。

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dispatch一个action，只要是同一个action type，在各个reducer中都可以接收到，各自修改各自的state

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看了你的問題直覺能想到的大方向有兩種

1. 將 clock 放到 todos 底下但這種方式長遠來說會越來越複雜

2. 另一個方式就是靠 thunk

// This is only for clock
const clockInitialState = {
  second: 5,
  minute: 0,
  runningType: 0 // 0: initial, 1: running, 2: done
}

function clock (state = clockInitialState, action) {
  let allSecond = state.minute * 60 + state.second
  let currentSecond = allSecond - 1
  let minute = Math.floor(currentSecond / 60);
  let second = currentSecond % 60
  let runningType = currentSecond === 0 ? 2 : 1;
  
  switch (action.type) {
    case COUNT_DOWN:
      return {
        ...state,
        minute: minute,
        second: second,
        runningType: runningType
      }
    case COUNT_RESET:
      return clockInitialState
    default:
      return state
  }
}

export const COUNT_DOWN = 'COUNT_DOWN'
export const COUNT_RESET = 'COUNT_RESET'

export function countDown() {
  return {
    type: COUNT_DOWN
  }
}

export function countReset() {
  return {
    type: COUNT_RESET
  }
}

// thunk
export function runCountDown() {
  return (dispatch, getState) => {
    let p = new Promise(function(resolve, reject) {
      let timer = setInterval(() => {
        console.log('!!');
        let { clock } = getState()
        if (clock.runningType === 2) {
          clearInterval(timer)
          resolve()
        } else {
          dispatch(countDown())
        }
      }, 1000)
    })
    return p.then(() => dispatch(addTodo('A')))
            .then(() => dispatch(addTodo('B')))
  }
}

我不確定這是最好的辦法但以我現階段的知識只能想出這個解法

完整範例