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redux常見(jiàn)問(wèn)題答疑

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摘要:為什么使用的核心是將組件化,由數(shù)據(jù)驅(qū)動(dòng)的展現(xiàn)。僅僅使用進(jìn)行開(kāi)發(fā)的痛點(diǎn)組件嵌套層級(jí)深,回調(diào)地獄。遵守容器組件與展示組件分離的原則。

為什么使用redux

React的核心是將UI組件化,由數(shù)據(jù)驅(qū)動(dòng)UI的展現(xiàn)。但是如何管理數(shù)據(jù)模型、組件與數(shù)據(jù)模型之間的通信,react并沒(méi)有很好的解決方案。Redux由flux演變而來(lái),同時(shí)簡(jiǎn)化了Flux的流程。

僅僅使用react進(jìn)行開(kāi)發(fā)的痛點(diǎn)

組件嵌套層級(jí)深,回調(diào)地獄。你可能會(huì)在一個(gè)較深層次的組件里需要更新全局state的某個(gè)字段,卻無(wú)奈只能通過(guò)從頂層組件一層一層傳遞下來(lái)的props進(jìn)行回調(diào)。

頁(yè)面的state不可預(yù)測(cè)。由于state缺乏一種可預(yù)測(cè)的機(jī)制,導(dǎo)致用戶在進(jìn)行一些頁(yè)面操作(更改state),或者異步請(qǐng)求有新的數(shù)據(jù)從服務(wù)端返回的時(shí)候,state的變化已經(jīng)不可控制,很容易產(chǎn)生bug

redux是怎么解決痛點(diǎn)的?

1、通過(guò)react-redux提供的Provider組件,在根組件外面包一層,這樣根組件,以及所有的子組件都能拿到store。實(shí)現(xiàn)的原理是基于React自身提供的context屬性,但是react官方不推薦直接在組件中使用this.context。所以react-redux提供了另一種方法connect,通過(guò)connect將普通的UI組件升級(jí)為容器組件,同時(shí)將獲取store的細(xì)節(jié)也一并封裝在生成容器組件的代碼中,從而容器組件可以直接拿到store





// 定義App組件
class Foo extends React.Component{
    render() {
    const { text } = this.props;
    return 
{text}
; } } const App = connect( mapStateToProps, mapDispatchToProps )(Foo);

2、使用純函數(shù)修改state,保證state變化可預(yù)測(cè)。每次更改都返回一個(gè)全新的state。
3、遵守容器組件與展示組件分離的原則。這是redux一個(gè)重要的思想,容器組件和展示組件各司其職。

容器組件

容器組件負(fù)責(zé)以下幾件事情:

處理數(shù)據(jù)邏輯

將store中的state轉(zhuǎn)變?yōu)閜rops傳遞給展示組件,對(duì)應(yīng)mapStateToProps

將注入了dispatch的函數(shù)作為props傳遞給展示組件,對(duì)應(yīng)mapDispatchToProps

有狀態(tài)的(展示組件則是無(wú)狀態(tài)的)

發(fā)起action,更新state

沒(méi)有DOM標(biāo)簽,沒(méi)有樣式

展示組件

可復(fù)用的組件,又稱為“傻瓜組件”,僅僅通過(guò)容器組件傳進(jìn)來(lái)的props進(jìn)行UI展示,以及操作回調(diào)。感知不到redux的存在,脫離redux框架也能使用,盡量保持無(wú)狀態(tài)(可包含少量的UI狀態(tài)),有以下幾條原則:

不理解數(shù)據(jù)邏輯:不關(guān)心數(shù)據(jù)是如何得到的,也不關(guān)心數(shù)據(jù)是如何改變的

只通過(guò)props獲取數(shù)據(jù)和操作回調(diào)

基本是無(wú)狀態(tài)的:必須有的話,可以是UI的狀態(tài)

什么時(shí)候引入容器組件

在引入之前,你可以先只用展示組件來(lái)構(gòu)建App。當(dāng)你發(fā)現(xiàn)有些組件并不會(huì)用到傳進(jìn)來(lái)的屬性,而僅僅只是把它往下傳給子組件的時(shí)候(并且層級(jí)比較深,超過(guò)兩層),你就要考慮引入容器組件了。

為什么使用immutable.js

首先,在沒(méi)有immutable.js的情況下,碰到較深層次的數(shù)據(jù)結(jié)構(gòu)時(shí),更新state會(huì)變得很麻煩。

舊的state:

{
  priceInfo: {
    price: 200,
    promotion: {
      offValue: 30
    }
  }
}

現(xiàn)在需要只更新offValue的值為50,該怎么處理呢?

let newState = _.cloneDeep(this.state);
newState.priceInfo.promotion.offValue = 50;
this.setState(newState);

可以看到,需要先深度復(fù)制一份this.state,然后修改offValue的值,最后執(zhí)行setState。除了這個(gè)過(guò)程的有點(diǎn)復(fù)雜,另外深度復(fù)制對(duì)象是挺耗性能的一件事。而通過(guò)Immutable.js,我們只需要這樣做:

let initialState = Immutable.fromJS({
  priceInfo: {
    price: 200,
    promotion: {
      offValue: 30
    }
  }
});
this.setState(initialState.setIn(["priceInfo", "promotion", "offValue"]), 50);
immutable.js帶來(lái)的性能提升

React做性能優(yōu)化時(shí)經(jīng)常用到 shouldComponentUpdate,只有深度比較才能得出正確的值,決定是否render,而深度比較是十分消耗性能的。Immutable 則提供了簡(jiǎn)潔高效的判斷數(shù)據(jù)是否變化的方法,只需 ===is 比較就能知道是否需要執(zhí)行render(),而這個(gè)操作幾乎零成本,所以可以極大提高性能。

immutable.js的數(shù)據(jù)結(jié)構(gòu)如何正確使用...擴(kuò)展運(yùn)算符

在jsx的語(yǔ)法中,你已經(jīng)習(xí)慣了使用擴(kuò)展運(yùn)算符來(lái)傳遞props,如下所示:

const titleInfo = {
  title: "主標(biāo)題",
  subTitle: "副標(biāo)題"
};

...

import Title from "title";

render() {
  return (
    
  );
}
</pre>
<p>但是如果你要傳遞的數(shù)據(jù)是<b>Immutable.Map</b>類型的,使用<b>...</b>擴(kuò)展運(yùn)算符,得到的數(shù)據(jù)可能與你期望的不太一樣,因?yàn)?b>Immutable.Map</b>的實(shí)例對(duì)象并不是plain object。<b>{...titleInfo}</b>等同于<b>Object.assign({}, titleInfo)</b>。Immutable.js提供了一種方法,可以實(shí)現(xiàn)這種轉(zhuǎn)換,如果是Map實(shí)例,只需調(diào)用<b>toObject()</b>(不要調(diào)用深度復(fù)制的<b>toJS</b>),就可以將Immutable.js的Map對(duì)象轉(zhuǎn)變?yōu)榭梢允褂?b>...</b>擴(kuò)展運(yùn)算符的plain object。</p>           
               
                                           
                       
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