我的reducer返回undefined，就像没有将其添加到CombineReducers一样，但是

Question

一切似乎都还不错，我已经调试了一段时间，但似乎不明白问题出在哪里

reducers / index.js

// @flow
import { combineReducers } from "redux";
import blocks from "./blocks";
import users from "./users";
import add from "./add";
const rootReducer = combineReducers({
    blocks,
    users,
    add
});
export default rootReducer;

reducers / add.js

// @flow
import AnyAction from "redux";
import * as ActionTypes from "./../constants";

const initialState = {
    number: 0
};

export default function Add(state = initialState, action) {
    switch (action.type) {
        case ActionTypes.ADD:
            return Object.assign({}, state, {
                number: number + 1
            });
        default:
            return state;
    }
}

连接功能

const { navigation, blocks, dispatch, number } = this.props;
.
.
.
const mapStateToProps = state => ({
    blocks: state.blocks,
    number: state.number
});

export default connect(mapStateToProps)(BlockList);

Answer 1

在合并化简器之后，您需要使用合并的化简器名称在mapStateToProps中访问化简器的状态。代替：

const mapStateToProps = state => ({
   blocks: state.blocks,
   number: state.number
});

具有此：

const mapStateToProps = state => ({
  blocks: state.blocks,
  number: state.add.number
});

现在，使用state.add.number，您可以访问add.js归约器中的数字状态。

我试图在这里重新创建您的问题：