用户单击弹出窗口中的按钮后关闭弹出窗口 react-leaflet [英] close popup react-leaflet after user click on button in popup

问题描述

所以基本上想为 react-leaflet Popup 组件自定义关闭，接缝这与原生 API 传单无关，但使用 react-leaflet 中的 react 组件我找不到解决方案.

So basically want to make custom close for react-leaflet Popup component, seams that is not a big problem to do with native API leaflet but with react component from react-leaflet I can't find the solution.

推荐答案

目前，我发现关闭弹出窗口的唯一方法如下:

at the moment, the only way I found to close the popup is the following:

constructor(props){
    super(props);
    this.popup = React.createRef();
}

// the magic
closePopusOnClick(){
    this.popup.current.leafletElement.options.leaflet.map.closePopup();
}

render(){
    return <Marker position={[this.props.lat, this.props.lng]}>
        <Popup ref={this.popup}>
            <Button onClick={this.closePopusOnClick}>Close popup</Button>
        </Popup>
    </Marker>;
}

希望有帮助！

这篇关于用户单击弹出窗口中的按钮后关闭弹出窗口 react-leaflet的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！