在地图上显示多个标记时,如何在单击标记时只打开一个信息window?
When displaying multiple markers on a map, how to open just one info window, when clicking on a marker?
我正在使用 react-google-maps 显示带有标记的地图,当您单击标记时,所有信息 windows 都会打开。我想在单击它时只显示一个标记的信息 window,而其他标记保持关闭状态。
这是我的代码:
<GoogleMap
defaultZoom={15}
defaultCenter={{ lat: 51.508530, lng: -0.076132 }}
>
{props.places && props.places.map((place, i) =>
<Marker onClick={props.onToggleOpen} key={i} position={{ lat: place.geometry.location.lat(), lng: place.geometry.location.lng() }} >
{props.isOpen && <InfoWindow onCloseClick={props.onToggleOpen}>
<div>{place.name}</div>
</InfoWindow>}
</Marker>
)}
</GoogleMap>
我用这个打开和关闭信息窗口
import { compose, withProps, withStateHandlers, withHandlers, withState } from "recompose";
...
withStateHandlers(() => ({
isOpen: false,
}), {
onToggleOpen: ({ isOpen, id }) => () => ({
isOpen: !isOpen,
})
}),
我正在绘制所有标记,并将它们显示在地图上。我怎样才能点击只打开一个标记信息窗口?
Here is a related question,但它不是用 React 制作的,也不使用 react-google-maps。
这更像是一个 React
问题。您可以将已点击的 Marker
的索引传递给 onToggleOpen
,而不是 isOpen
,您可以使用 selectedPlace
状态来保存已点击的 Marker
的索引并使用此索引渲染正确 InfoWindow
.
这里有一个例子(没有完全测试,但你可以理解):
/*global google*/
import React from "react"
import { compose, withProps, withHandlers, withState, withStateHandlers } from "recompose"
import { withScriptjs, withGoogleMap, GoogleMap, Marker, InfoWindow } from "react-google-maps"
const MyMapComponent = compose(
withProps({
googleMapURL: "https://maps.googleapis.com/maps/api/js?v=3.exp&libraries=geometry,drawing,places",
loadingElement: <div style={{ height: `100%` }} />,
containerElement: <div style={{ height: `400px` }} />,
mapElement: <div style={{ height: `100%` }} />,
}),
withScriptjs,
withGoogleMap,
withState('places', 'updatePlaces', ''),
withState('selectedPlace', 'updateSelectedPlace', null),
withHandlers(() => {
const refs = {
map: undefined,
}
return {
onMapMounted: () => ref => {
refs.map = ref
},
fetchPlaces: ({ updatePlaces }) => {
let places;
const bounds = refs.map.getBounds();
const service = new google.maps.places.PlacesService(refs.map.context.__SECRET_MAP_DO_NOT_USE_OR_YOU_WILL_BE_FIRED);
const request = {
bounds: bounds,
type: ['hotel']
};
service.nearbySearch(request, (results, status) => {
if (status == google.maps.places.PlacesServiceStatus.OK) {
console.log(results);
updatePlaces(results);
}
})
},
onToggleOpen: ({ updateSelectedPlace }) => key => {
updateSelectedPlace(key);
}
}
}),
)((props) => {
console.log(props);
return (
<GoogleMap
onTilesLoaded={props.fetchPlaces}
ref={props.onMapMounted}
onBoundsChanged={props.fetchPlaces}
defaultZoom={15}
defaultCenter={{ lat: 51.508530, lng: -0.076132 }}
>
{props.places && props.places.map((place, i) =>
<Marker onClick={() => props.onToggleOpen(i)} key={i} position={{ lat: place.geometry.location.lat(), lng: place.geometry.location.lng() }}>
{props.selectedPlace === i && <InfoWindow onCloseClick={props.onToggleOpen}>
<div>
{props.places[props.selectedPlace].name}
</div>
</InfoWindow>}
</Marker>
)}
</GoogleMap>
)
})
export default class MyFancyComponent extends React.PureComponent {
render() {
return (
<MyMapComponent />
)
}
}
我正在使用 react-google-maps 显示带有标记的地图,当您单击标记时,所有信息 windows 都会打开。我想在单击它时只显示一个标记的信息 window,而其他标记保持关闭状态。
这是我的代码:
<GoogleMap
defaultZoom={15}
defaultCenter={{ lat: 51.508530, lng: -0.076132 }}
>
{props.places && props.places.map((place, i) =>
<Marker onClick={props.onToggleOpen} key={i} position={{ lat: place.geometry.location.lat(), lng: place.geometry.location.lng() }} >
{props.isOpen && <InfoWindow onCloseClick={props.onToggleOpen}>
<div>{place.name}</div>
</InfoWindow>}
</Marker>
)}
</GoogleMap>
我用这个打开和关闭信息窗口
import { compose, withProps, withStateHandlers, withHandlers, withState } from "recompose";
...
withStateHandlers(() => ({
isOpen: false,
}), {
onToggleOpen: ({ isOpen, id }) => () => ({
isOpen: !isOpen,
})
}),
我正在绘制所有标记,并将它们显示在地图上。我怎样才能点击只打开一个标记信息窗口? Here is a related question,但它不是用 React 制作的,也不使用 react-google-maps。
这更像是一个 React
问题。您可以将已点击的 Marker
的索引传递给 onToggleOpen
,而不是 isOpen
,您可以使用 selectedPlace
状态来保存已点击的 Marker
的索引并使用此索引渲染正确 InfoWindow
.
这里有一个例子(没有完全测试,但你可以理解):
/*global google*/
import React from "react"
import { compose, withProps, withHandlers, withState, withStateHandlers } from "recompose"
import { withScriptjs, withGoogleMap, GoogleMap, Marker, InfoWindow } from "react-google-maps"
const MyMapComponent = compose(
withProps({
googleMapURL: "https://maps.googleapis.com/maps/api/js?v=3.exp&libraries=geometry,drawing,places",
loadingElement: <div style={{ height: `100%` }} />,
containerElement: <div style={{ height: `400px` }} />,
mapElement: <div style={{ height: `100%` }} />,
}),
withScriptjs,
withGoogleMap,
withState('places', 'updatePlaces', ''),
withState('selectedPlace', 'updateSelectedPlace', null),
withHandlers(() => {
const refs = {
map: undefined,
}
return {
onMapMounted: () => ref => {
refs.map = ref
},
fetchPlaces: ({ updatePlaces }) => {
let places;
const bounds = refs.map.getBounds();
const service = new google.maps.places.PlacesService(refs.map.context.__SECRET_MAP_DO_NOT_USE_OR_YOU_WILL_BE_FIRED);
const request = {
bounds: bounds,
type: ['hotel']
};
service.nearbySearch(request, (results, status) => {
if (status == google.maps.places.PlacesServiceStatus.OK) {
console.log(results);
updatePlaces(results);
}
})
},
onToggleOpen: ({ updateSelectedPlace }) => key => {
updateSelectedPlace(key);
}
}
}),
)((props) => {
console.log(props);
return (
<GoogleMap
onTilesLoaded={props.fetchPlaces}
ref={props.onMapMounted}
onBoundsChanged={props.fetchPlaces}
defaultZoom={15}
defaultCenter={{ lat: 51.508530, lng: -0.076132 }}
>
{props.places && props.places.map((place, i) =>
<Marker onClick={() => props.onToggleOpen(i)} key={i} position={{ lat: place.geometry.location.lat(), lng: place.geometry.location.lng() }}>
{props.selectedPlace === i && <InfoWindow onCloseClick={props.onToggleOpen}>
<div>
{props.places[props.selectedPlace].name}
</div>
</InfoWindow>}
</Marker>
)}
</GoogleMap>
)
})
export default class MyFancyComponent extends React.PureComponent {
render() {
return (
<MyMapComponent />
)
}
}