2017-03-01 57 views
0

我有下面的反應組件,問題是它不會在檢測到onClick事件(觸發handleClick函數)時自動呈現。React state not rerendering

如果我寫這個.forceUpdate()比它的工作正常。這不應該是正常的,它應該沒有它的工作。

import React from 'react'; 
import Loading from './Loading'; 

import CameraStore from "../stores/CameraStore"; 
import * as CameraActions from "../actions/CameraActions"; 

class LocationCameras extends React.Component { 
    constructor(props){ 
     super(props); 

     this.state = { 
      updating: false 
     }; 

     this.handleClick = this.handleClick.bind(this) 
    } 

    handleSwitch(){ 

    } 

    handleClick(){ 

     if(this.state.updating == false){ 
      this.state.updating = true; 
     } else if (this.state.updating == true) { 
      this.state.updating = false; 
     } 

     console.log(this.state.updating); 
     this.forceUpdate(); 

    } 

    render(){ 
     var that = this; 
     return(
      <div className="cameras"> 
       <div className="panel panel-default"> 
        <div className="panel-heading"> 
         Client: {this.props.locationName} 
         <span className='camera--make-updates' onClick={this.handleClick} > 
          Update 
         </span> 
        </div> 
        <div className="panel-body"> 
         { 
          this.props.cameras.map(function(camera,index){ 

           if (camera.disabled !== "1"){ 
            var statusClassName; 
            var checkStatus; 
            if (camera.status == 1){ 
             var statusClassName = 'location--camera active'; 
             checkStatus = true; 
            } 
            else { 
             var statusClassName = 'location--camera disabled'; 
             checkStatus = false; 
            } 

            var handleSwitch = function(event){ 
             var cameraId = camera.cameraId; 
             that.handleSwitch(event, cameraId, index); 

            } 

              return (

               <div className={statusClassName} key={camera.cameraId}> 
                { 
                that.state.updating == true ? (

                 <div className="camera--switch"> 

                  <input type="checkbox" id={'switch' + camera.cameraId} defaultChecked={checkStatus} onChange={handleSwitch} className="switch--input"/> 
                  <label htmlFor={'switch' + camera.cameraId} className="switch--label"></label> 

                 </div> 

                 ) : (

                  <div className="camera--status"></div> 

                 ) 


                } 

                <div className="camera--name-id"> 
                { camera.name !== '' ? (
                  camera.name 
                 ) : (
                  '#' + camera.cameraId 
                 ) 
                } 



                </div> 

               </div> 
              ) 
             } 
           } 

           ) 

          } 

          </div> 
        </div> 
       </div> 
     ) 
    } 
} 

module.exports = LocationCameras; 

回答

2

使用this.setState({updating: false}) insted的this.state.updating = false;和組件將重新呈現;

要遵循@Felix評論,效果會更好更改功能handleClick到

handleClick(){ 
    this.setState({ updating: !this.state.updating }) 
} 
+2

更好:整個函數體可以用'this.setState更換({更新:this.state.updating})' –

+0

@Felix,當然,你是對的。我只是展示他如何調用重新渲染。 – Andrew

3

read the React documentation

使用國家正確

有三件事情你應該知道setState()

不要修改國家直接

例如,這將不會重新渲染的成分:

// Wrong 
this.state.comment = 'Hello'; 

相反,使用setState()

// Correct 
this.setState({comment: 'Hello'}); 

只有在那裏你可以把分配this.state是構造函數。

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