blocks|key|1175787|text|我不认为对于这个简单的范例，您需要经历设置redux的所有麻烦(我喜欢Redux，但有时我们不需要使用它)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175788|下面是如何使用新的context实现它。|offset|length|style|CODE|1175789|​|1175790|📷|atomic|1175791|1175792|它简单得多，使用起来也很好。检查一下，这里，如果不清楚，请告诉我。|blockquote|1175793|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/d24ed1d187fba0d9cda58f0f602bd41e.svg|imageAlt|1|LINK|MUTABLE|url|https://reactjs.org/docs/context.html#api^0|0|9|7|0|0|0|1|0|0|0|J|2|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|19|8|@]|9|@$D|1A|E|1B|1|1C]]|A|$]]|$1|M|3|I|5|6|7|1D|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1E|8|@]|9|@$D|1F|E|1G|1|1H]]|A|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X|Y|-4]]|Z|$5|10|U|11|A|$12|13]]]]

I don't think for this simple paradigm you'll need to go through all the hassle of setting redux up(I love Redux, but sometimes we don't have to use it)

Here's how you could implement it with the new <code>context</code>.

<a href="https://codesandbox.io/s/ko1njmo25?fontsize=14" rel="nofollow noreferrer"><img src="https://codesandbox.io/static/img/play-codesandbox.svg" alt="Edit React Hooks"></a>

<blockquote>
 It's farily simple and nice to use. Check it <a href="https://reactjs.org/docs/context.html#api" rel="nofollow noreferrer">here</a>
 Let me know if that isn't clear.
</blockquote>

blocks|key|2018592|text|登录应该是MyAppBar的子组件。然后，您可以将函数handleLogin作为登录的道具。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2018593|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Login should be a children component of MyAppBar. Then you could pass function handleLogin as a props to Login.

blocks|key|400358|text|你应该使用REDUX。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|400359|主要是因为它允许您拥有任何组件都可以读取或同步的状态。|400360|我很少使用react的状态/setState。|400361|在大多数情况下，redux状态很好地改变/观察。|400362|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

You should use REDUX. 

Mostly because it lets you have a state that any component can read or can be synced with.

I rarely use react's state/setState. 

In most cases redux state is good to change/watch.

blocks|key|2018610|text|在这样的场景中使用redux更好，因为这使得组件之间的通信非常容易。否则，如果您想走艰苦的道路，那么创建一个父组件，它将包装登录和MyAppBar。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2018611|++class+ParentComp+extends+Component{
++++++constructor(props){+
+++++++++super(props);
+++++++++this.state={
+++++++++++++isLoggedIn:false
+++++++++};
++++++++this.setLoginState=this.setLoginState.bind(this);
++++++}
++++++setLoginState(value){
++++++++++this.setState({isLoggedIn:value});
++++++}

+++++++render(){
+++++++++return(<div>
+++++++++++++++++++<MyAppBar+isLoggedIn={this.state.isLoggedIn}+/>
+++++++++++++++<Route+path={path}+render={props=><Login+{..props}+setLoginState={this.setLoginState}/>}/>
++++++++++++++++</div>);
+++++++}
++}|code-block|syntax|javascript|2018612|在LoginComponent中调用setLoginState方法|2018613|+++this.props.setLoginState(true);|2018614|在myappbar组件中，使用isLoggedIn支柱有条件地呈现|2018615|+++renderLogin(){
+++++if(this.props.isLoggedIn){+
++++++++++//+return+loggedin+jsx
+++++}else{
+++++++++return+the+alternative
+++++}
+++}|2018616|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

its better to use redux in such scenarios, as that makes communication between components very easy. Else if you want to go the hard way, then create a parent component, that will wrap both login and MyAppBar.

<pre><code> class ParentComp extends Component{
 constructor(props){ 
 super(props);
 this.state={
 isLoggedIn:false
 };
 this.setLoginState=this.setLoginState.bind(this);
 }
 setLoginState(value){
 this.setState({isLoggedIn:value});
 }

 render(){
 return(&lt;div&gt;
 &lt;MyAppBar isLoggedIn={this.state.isLoggedIn} /&gt;
 &lt;Route path={path} render={props=&gt;&lt;Login {..props} setLoginState={this.setLoginState}/&gt;}/&gt;
 &lt;/div&gt;);
 }
 }
</code></pre>

in your LoginComponent call the setLoginState method

<pre><code> this.props.setLoginState(true);
</code></pre>

and in your myappbar component
 use the isLoggedIn prop to render conditionally

<pre><code> renderLogin(){
 if(this.props.isLoggedIn){ 
 // return loggedin jsx
 }else{
 return the alternative
 }
 }
</code></pre>

I have the following structure in index.js :

<pre><code>&lt;MyAppBar // + properties/&gt;
 &lt;Route path={path} component={Login}/&gt;
</code></pre>

MyAppBar.js

<pre><code> class MyAppBar extends React.Component {

state = {
 auth: false,
 anchorElement: null
};

handleLogin(){
 this.setState({auth:true});
}

//another code
//end component

export default withStyles(styles)(MyAppBar)
</code></pre>

I want to update MyAppBar state auth flag to be true from within the Login component action handler (i.e. handleLogin), but I do not know how.

I've tried with importing MyAppBar (all component) on the Login page, but it says that 

<pre><code>MyAppBar.handleLogin(); 
</code></pre>

is not a function... Thank you!

Thank you!

Set state from another component React js

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在index.js中有以下结构：<MyAppBar // + properties/> <Route path={path} component={Login}/>MyAppBar.js class MyAppBar extends React.Component {state = { auth: false, anchorElement: null};handleLogin(){ thi

问来自另一个组件的Set状态响应js。
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