我正在使用 Redux,并希望动态包含目录中的所有文件。
/redux/index.js
// Actions
import * as authActions from './auth/authActions';
import * as deviceActions from './device/deviceActions';
import * as globalActions from './global/globalActions';
import * as menuActions from './menu/menuActions';
... etc
export const actions = [
authActions,
deviceActions,
globalActions,
menuActions,
...
];
// Reducers
import auth from './auth/authReducer';
import device from './device/deviceReducer';
import global from './global/globalReducer';
import menu from './menu/menuReducer';
...
import { combineReducers } from 'redux';
export const rootReducer = combineReducers({
auth,
device,
global,
menu,
...
});
在上面(简化的)示例中,所有文件的结构如下:
/redux/
/auth/
authActions.js
authReducer.js
/device/
deviceActions.js
deviceReducer.js
/global/
globalActions.js
globalReducer.js
/menu/
menuActions.js
menuReducer.js
...
在这个index.js文件中,如果我可以动态读取redux目录中的所有目录,并动态要求导出actions和reducers,那么维护起来会容易得多。
在常规节点环境中,我会执行类似的操作(未测试,但说明了示例):
import fs from 'fs'
import path from 'path'
import { combineReducers } from 'redux'
let actions = []
let reducers {}
fs
.readdirSync(__dirname).filter((file) => {
// Only directories
return fs.statSync(path.join(__dirname, file)).isDirectory();
})
.forEach((module) => {
const moduleActions = require(path.join(__dirname, module, `${module}Actions.js`);
const moduleReducer = require(path.join(__dirname, module, `${module}Reducer.js`);
actions.push(moduleActions)
reducers[module] = moduleReducer.default
});
export actions
export const rootReducer = combineReducers(reducers)
问题是fs
module 不是react-native 中能够动态迭代代码库中的目录的东西。有react-native-fs,但那是为了实际访问设备上的文件系统(在编译应用程序之后)[我认为?]。上面的内容比单独要求所有操作和化简器并在操作数组和化简器对象中指定它们要干净得多。
有任何想法吗?