ReactJS：警告：setState（...）：无法在现有状态转换期间更新

我正在尝试从渲染视图重构以下代码：

<Button href="#" active={!this.state.singleJourney} onClick={this.handleButtonChange.bind(this,false)} >Retour</Button>

绑定在构造函数内的版本。原因是渲染视图中的绑定会给我带来性能问题，尤其是在低端手机上。

我创建了以下代码，但我不断收到以下错误（很多）。看起来应用程序陷入了循环：

Warning: setState(...): Cannot update during an existing state transition (such as within `render` or another component's constructor). Render methods should be a pure function of props and state; constructor side-effects are an anti-pattern, but can be moved to `componentWillMount`.

下面是我使用的代码：

var React = require('react');
var ButtonGroup = require('react-bootstrap/lib/ButtonGroup');
var Button = require('react-bootstrap/lib/Button');
var Form = require('react-bootstrap/lib/Form');
var FormGroup = require('react-bootstrap/lib/FormGroup');
var Well = require('react-bootstrap/lib/Well');

export default class Search extends React.Component {

    constructor() {
        super();

        this.state = {
            singleJourney: false
        };

        this.handleButtonChange = this.handleButtonChange.bind(this);
    }

    handleButtonChange(value) {
        this.setState({
            singleJourney: value
        });
    }

    render() {

        return (
            <Form>

                <Well style={wellStyle}>

                    <FormGroup className="text-center">

                        <ButtonGroup>
                            <Button href="#" active={!this.state.singleJourney} onClick={this.handleButtonChange(false)} >Retour</Button>
                            <Button href="#" active={this.state.singleJourney} onClick={this.handleButtonChange(true)} >Single Journey</Button>
                        </ButtonGroup>
                    </FormGroup>

                </Well>

            </Form>
        );
    }
}

module.exports = Search;

看来您不小心调用了handleButtonChange方法在你的渲染方法中，你可能想要做onClick={() => this.handleButtonChange(false)}反而。

如果您不想在 onClick 处理程序中创建 lambda，我认为您需要有两个绑定方法，每个参数一个。

In the constructor:

this.handleButtonChangeRetour = this.handleButtonChange.bind(this, true);
this.handleButtonChangeSingle = this.handleButtonChange.bind(this, false);

并且在render method:

<Button href="#" active={!this.state.singleJourney} onClick={this.handleButtonChangeSingle} >Retour</Button>
<Button href="#" active={this.state.singleJourney} onClick={this.handleButtonChangeRetour}>Single Journey</Button>