React - 当父状态组件发生更改时,如何强制子组件重新呈现？

Question

在我的笔http://codepen.io/seunlanlege/pen/PbYNor?editors=0010#0

发生的事情非常简单.this.send()调用方法时 this.state.chat被发送到子组件<Messages /> which is then concatenated to thethis.state.messages`数组并呈现.

但问题是,渲染后更新1秒,this.state.sent这是一个支持<Messages />但消息组件不重新渲染.

请问如何强制它重新渲染？

class Messages extends React.Component{
constructor(props){
super(props);
}
render(){
return(
`<p id={this.props.key} className={this.props.style}>{this.props.msg}<span>{this.props.sent}</span></p>`);
}
}


class Chat extends React.Component {
constructor(props) {
super(props);
this.count = 0;
    this.state = {chat :'',messages:[],sent:[]}; 
}
chatChange(event){
    this.setState({chat : event.target.value});
}
send(e){
    e.preventDefault();
    let x = this.state.messages;
    this.setState({messages:x.concat([<Messages style="me" msg={this.state.chat} key={this.state.chat} sent={(this.state.sent[this.count])?this.state.sent[this.count]:''} />])});
this.setState({chat:''});
setTimeout(()=>{
  this.setState({sent:this.state.sent.concat(['sent'])});
  this.count++;
},1000);
}
render () {
    return(
        `<div>
        <div id="top"><h1>Chat With Customer Support</h1></div>
        <div id="chatbox">
        {this.state.messages}
        </div>
        <div id="bottom">
        <form onSubmit={this.send.bind(this)}>
        <span>
        <input value={this.state.chat} type="text" onChange={this.chatChange.bind(this)} id="chat" value={this.state.chat} />
        <button type="submit" id="send">Send</button>
        </span>
        </form>
        </div>
        </div>`
        );
}
}ReactDOM.render(<Chat name="seun" />,document.getElementById('app')):</code>

Answer 1

你的codepen很好,只有你错过的是消息组件的正确密钥,

为了正确地渲染组件,每个组件应该有不同的密钥,所以我改变了一段代码

<Messages style="me" msg={this.state.chat} key={this.state.chat} sent={(this.state.sent[this.count])?this.state.sent[this.count]:''} />

至

<Messages style="me" msg={this.state.chat} key={this.count} sent={(this.state.sent[this.count])?this.state.sent[this.count]:''} />

和代码笔工作正常

编辑这解决了问题

而不是将组件保存到状态,动态构建它们.检查这个codepen.io/abhirathore2006/pen/JbjjPx