React:子项onclick更改父级的状态以进行重新渲染

Question

我真的很反应.我正在尝试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,它应该向父组件发送"true"或"false",以便它知道是否要呈现它.我认为应该用这样的道具来完成:

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

但我不能让它工作.你能给出任何提示吗？这是完整的父母 - 孩子

parent - 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;

Answer 1

你需要做的是将一个方法从父方法传递给子节点,它可以在单击按钮时调用.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})
}

将此方法传递给您的子组件
<Mainpage_Addscreen buttonClick={this.changeButtonState.bind(this)} />

最后在子组件中,
<Button .... onClick={this.props.buttonClick} />

Answer 2

您可能需要的是一个回调函数，父母将其作为道具传递给您的孩子，然后您的孩子可以进行呼叫。

像这样：

// In parent
constructor() {
  ...
  this.onChildClicked = this.onChildClicked.bind(this);
}

onChildClicked() {
  this.setState({childWasClicked : True });
}

render() {
  ...
  return (
    <div>
      <Child onClicked={this.onChildClicked} />
    </div>
  )
}

// In child
render() {
  ...
  return (
    <div>
      <button onClick={this.props.onClicked} />
    </div>
  )
}

PS：我注意到您有一个大写的<Button>部分。这通常用于您自己定义的组件。标准DOM元素采用小写字母，例如<div>，<p>等等