开发中有从详情页返回列表页的需求,这样一来页面返回后使用react-router会直接刷新页面,导致页面中的分页和搜索条件全部丢失,用户体验不佳,所以就必须将列表页的状态进行缓存。
网上搜索大概有几种方法:
1、使用localStorage/sessionStorage进行页面的状态的保存,跳转页面后再进行获取,这种方法虽然可行,但是从根本来说还是从新向后台再一次请求了数据,不算最佳方案。
2、react-activation,尝试未果
3、react-kepper,需要将项目的react-router替换掉,风险较大,慎用
4、react-router-cache-route,简单易用,最佳方案
基本使用方法:
就是把Switch替换成CacheSwitch,因为因为Switch组件只保留第一个匹配状态的路由,卸载掉其他路由
把Route替换成CacheRoute
注意:详情页面回退列表页时使用this.props.history.push(‘’路径')可以实现页面的缓存
但当使用this.props.history.go(-1)则缓存失败
具体代码如下:
import React,{Component} from 'react'
import { Route} from 'react-router-dom'
import {CacheRoute,CacheSwitch} from 'react-router-cache-route'
import PublishExpend from './publish/publishExpend.js'
import PublishExpendDetail from './publish/publishExpendDetail.js'
import EditorExpend2018 from './editor/editorExpend2018.js'
import EditorExpend2019 from './editor/editorExpend2019.js'
import EditorExpend2020 from './editor/editorExpend2020.js'
import ExpenseSearch from './search/expenseSearch.js'
class Expense extends Component{
render(){//:productId?,后面加上?的意思是可有可无,有的话匹配到有id的路由,没有id的话就匹配之前的路由
return(
<div className='Expense'>
<CacheSwitch>
<CacheRoute exact path='/platform/expense/publishExpend' component={PublishExpend}/>
<Route path='/platform/expense/publishExpend/detail/:expendId?' component={PublishExpendDetail}/>
<Route path='/platform/expense/editorExpend2018' component={EditorExpend2018}/>
<Route path='/platform/expense/editorExpend2019' component={EditorExpend2019}/>
<Route path='/platform/expense/editorExpend2020' component={EditorExpend2020}/>
<Route exact path='/platform/expense/expenseSearch' component={ExpenseSearch}/>
</CacheSwitch>
</div>
)
}
}
export default Expense;
其他方法了解可以前往:https://github.com/CJY0208/react-router-cache-route/blob/HEAD/README_CN.md