我正在使用immutable.JS通过redux-immutablejs来管理我的商店。我现在想使用redux-form库,但我在组合reducers时遇到了问题。
Redux-immutable提供了一个combineReducers函数,该函数将检查传递给它的所有reducers是否都返回不可变对象。
Redux本身提供了一个不执行此类检查的combineReducers函数。
Redux-form要求您包含它们的缩减程序,但我不能使用Redux immutable的combineReducers执行此操作,因为它将失败。
所以我尝试做的基本上是将这两个函数的输出结合起来,如下所示:
import { combineReducers } from 'redux';
import { combineReducers as combineReducersUtils } from 'redux-utils';
import {reducer as formReducer} from 'redux-form';
const mainReducers = combineReducersUtils({
devices, alarms
});
const extraReducers = combineReducers({
form: formReducer
});
export default (mainReducers + extraReducers);最后一行显然不起作用,但基本上说明了我想要的是什么。
感谢您抽出时间阅读这篇文章。
发布于 2015-11-28 05:47:23
也许是这样的?
function rootReducer(state, action) {
const newState = combineReducersUtils({devices, alarms})(state, action);
return combineReducers({form: formReducer})(newState, action);
}它应该可以工作,因为combineReducers的返回值只是一个减法器:
(函数):一个reducer,它调用reducers对象中的每个reducer,并构造一个具有相同形状的状态对象。
https://github.com/rackt/redux/blob/master/docs/api/combineReducers.md
已更新,不会在每个派单上创建新函数:
const mainReducers = combineReducersUtils({devices, alarms});
const formReducers = combineReducers({form: formReducer});
function rootReducer(state, action) {
return formReducers(mainReducers(state, action), action);
}https://stackoverflow.com/questions/33720361
复制相似问题