反应:孩子点击以更改父母的状态以进行重新渲染

问题描述 投票:8回答:2

我真的很陌生。我试图让父母根据状态更改显示的页面。我的子组件中有一个按钮,应该向父组件发送“ true”或“ false”,以便它知道是否进行渲染。我认为应该用这样的道具来完成:

this.state = {
   btnNewScreen: this.props.btnNewScreen //true or false
};

但我无法使其正常工作。你能给点建议吗?这是完整的父母-子女

父-maindisplay.js

import React from 'react';
import Mainpage_Addscreen from '../components/subcomponents/mainpage-addscreen';
import Mainpage_Showscreens from '../components/subcomponents/mainpage-showscreens';
//
class MainDisplay extends React.Component {
    constructor() {
        super();
        this.state = {
            btnNewScreen: false //should be this.props.btnNewScreen?
        };
    }

    render() {
        var renderThis;
        if (!this.state.btnNewScreen) {
            renderThis =
                <div>
                    <Mainpage_Addscreen />
                    <Mainpage_Showscreens />
                </div>
        }
        else {
            //renderThis = <AddScreen />
            renderThis =
                <div>
                    <Mainpage_Addscreen />
                    <h3>Change to this when true (button click)</h3>
                </div>
        }
        return (
            <div>
                {renderThis}
            </div>
        );
      }
    }


    export default MainDisplay;

child-mainpage-addscreen.js

import React from 'react';

import Glyphicon from 'react-bootstrap/lib/Glyphicon';
import Button from 'react-bootstrap/lib/Button';

class Mainpage_Addscreen extends React.Component {
    constructor() {
        super();
        this.state = {
            btnNewScreen: true
        };
        this.newScreen = this.newScreen.bind(this);
    }

    newScreen(e) {
        this.setState({ btnNewScreen: !this.state.btnNewScreen });
        console.log(this.state.btnNewScreen);
    }
    render() {
        var text = this.state.btnNewScreen ? 'Add new' : 'Screens';
        return (
            <div className="main_window col-sm-offset-1 col-sm-10">
                <h3 id="addscreens">Screens: </h3>
                <Button id="addScreen" className="btn btn-primary dropdown-toggle" type="button" data-toggle="dropdown" onClick={this.newScreen}><Glyphicon id="refresh_screens" glyph="plus" />&nbsp; {text}</Button>
            </div>
        );
    }
}


export default Mainpage_Addscreen;
reactjs components
2个回答
22
投票

您需要做的是将方法从父级传递给子级,单击该按钮时可以调用该方法。属于父级的此方法将更改状态。在MainPage.js中

changeButtonState(event) {
    this.setState({btnNewScreen: !this.state.btnNewScreen})
}

将此方法作为]传递给您的子组件>

<Mainpage_Addscreen buttonClick={this.changeButtonState.bind(this)} />

最后是子组件,

<Button ....   onClick={this.props.buttonClick} />

5
投票

您可能需要一个回调函数,父母将其作为道具传递给孩子,然后孩子可以呼叫。

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