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[英]React JS change component style on changing another component's state
[英]Change a react component's state on a click of another component
我试图根据其状态显示/隐藏一个组件,并且我想在单击第三个组件时对其进行更改。
//导航栏
export class NavigationBar extends React.Component {
constructor(props) {
super(props);
this.state = {
showNotification: false,
}
}
handleNotification = () => this.setState({
showNotification: !this.state.showNotification,
});
{ this.state.showNotification ? <Outside><Notifications /></Outside> : null}
//外部组件,负责检测外部是否发生了点击。
export default class Outside extends Component {
constructor(props) {
super(props);
this.setWrapperRef = this.setWrapperRef.bind(this);
this.handleClickOutside = this.handleClickOutside.bind(this);
}
componentDidMount() {
document.addEventListener('mousedown', this.handleClickOutside)
}
setWrapperRef(node) {
this.wrapperRef = node;
}
handleClickOutside(event) {
if(this.wrapperRef && !this.wrapperRef.contains(event.target)) {
console.log("clicked outside notifications");
this.setState({
showNotification: false
})
}
}
render() {
return (
<div ref={this.setWrapperRef}>
{this.props.children}
</div>
)
}
}
Outside.propTypes = {
children: PropTypes.element.isRequired
}
我的疑问是如何根据在Outside component
内部检测到的事件来更改导航栏中的状态?
在父级中,您需要向下到事件处理程序之外:
<Outside toggleNofitication={this.handleNotification}><Notifications /></Outside>
在外部,只需在事件触发时调用toggleNofitication
:
handleClickOutside = () => {
// ...
this.props.toggleNofitication()
}
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