使用Google Maps API v3中的多个标记自动对中地图

2020/10/12 13:01 · javascript ·  · 0评论

这是我用来显示具有3个图钉/标记的地图的方法:

<script>
  function initialize() {
    var locations = [
      ['DESCRIPTION', 41.926979, 12.517385, 3],
      ['DESCRIPTION', 41.914873, 12.506486, 2],
      ['DESCRIPTION', 41.918574, 12.507201, 1]
    ];

    var map = new google.maps.Map(document.getElementById('map'), {
      zoom: 15,
      center: new google.maps.LatLng(41.923, 12.513),
      mapTypeId: google.maps.MapTypeId.ROADMAP
    });

    var infowindow = new google.maps.InfoWindow();

    var marker, i;

    for (i = 0; i < locations.length; i++) {
      marker = new google.maps.Marker({
        position: new google.maps.LatLng(locations[i][1], locations[i][2]),
        map: map
      });

      google.maps.event.addListener(marker, 'click', (function(marker, i) {
        return function() {
          infowindow.setContent(locations[i][0]);
          infowindow.open(map, marker);
        }
      })(marker, i));
    }
  }

  function loadScript() {
    var script = document.createElement('script');
    script.type = 'text/javascript';
    script.src = 'https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false&' + 'callback=initialize';
    document.body.appendChild(script);
  }

  window.onload = loadScript;
</script>

<div id="map" style="width: 900px; height: 700px;"></div>

我正在寻找一种避免使用手动“找到”地图中心的方法center: new google.maps.LatLng(41.923, 12.513)有没有一种方法可以自动使地图以三个坐标为中心?

有一种更简单的方法,即扩展一个空白LatLngBounds而不是从两点显式地创建一个空白(有关更多详细信息,请参阅此问题

应该看起来像这样,添加到您的代码中:

//create empty LatLngBounds object
var bounds = new google.maps.LatLngBounds();
var infowindow = new google.maps.InfoWindow();    

for (i = 0; i < locations.length; i++) {  
  var marker = new google.maps.Marker({
    position: new google.maps.LatLng(locations[i][1], locations[i][2]),
    map: map
  });

  //extend the bounds to include each marker's position
  bounds.extend(marker.position);

  google.maps.event.addListener(marker, 'click', (function(marker, i) {
    return function() {
      infowindow.setContent(locations[i][0]);
      infowindow.open(map, marker);
    }
  })(marker, i));
}

//now fit the map to the newly inclusive bounds
map.fitBounds(bounds);

//(optional) restore the zoom level after the map is done scaling
var listener = google.maps.event.addListener(map, "idle", function () {
    map.setZoom(3);
    google.maps.event.removeListener(listener);
});

这样,您可以使用任意数量的点,而无需事先知道顺序。

演示jsFiddle在这里:http : //jsfiddle.net/x5R63/

我认为您必须计算纬度最小值和经度最小值:这是一个示例,该函数用于将点居中:

//Example values of min & max latlng values
var lat_min = 1.3049337;
var lat_max = 1.3053515;
var lng_min = 103.2103116;
var lng_max = 103.8400188;

map.setCenter(new google.maps.LatLng(
  ((lat_max + lat_min) / 2.0),
  ((lng_max + lng_min) / 2.0)
));
map.fitBounds(new google.maps.LatLngBounds(
  //bottom left
  new google.maps.LatLng(lat_min, lng_min),
  //top right
  new google.maps.LatLng(lat_max, lng_max)
));

要找到地图的确切中心,您需要将纬度/经度坐标转换为像素坐标,然后找到像素中心并将其转换回纬度/经度坐标。

您可能不会注意到或介意漂移,具体取决于您位于赤道以北或以南的距离。您可以通过在setInterval内执行map.setCenter(map.getBounds()。getCenter())来查看漂移,该漂移在接近赤道时将逐渐消失。

您可以使用以下内容在经度/纬度和像素坐标之间转换。像素坐标基于完全放大的整个平面,但是您可以找到其中心并将其切换回经/纬度。

   var HALF_WORLD_CIRCUMFERENCE = 268435456; // in pixels at zoom level 21
   var WORLD_RADIUS = HALF_WORLD_CIRCUMFERENCE / Math.PI;

   function _latToY ( lat ) {
      var sinLat = Math.sin( _toRadians( lat ) );
      return HALF_WORLD_CIRCUMFERENCE - WORLD_RADIUS * Math.log( ( 1 + sinLat ) / ( 1 - sinLat ) ) / 2;
   }

   function _lonToX ( lon ) {
      return HALF_WORLD_CIRCUMFERENCE + WORLD_RADIUS * _toRadians( lon );
   }

   function _xToLon ( x ) {
      return _toDegrees( ( x - HALF_WORLD_CIRCUMFERENCE ) / WORLD_RADIUS );
   }

   function _yToLat ( y ) {
      return _toDegrees( Math.PI / 2 - 2 * Math.atan( Math.exp( ( y - HALF_WORLD_CIRCUMFERENCE ) / WORLD_RADIUS ) ) );
   }

   function _toRadians ( degrees ) {
      return degrees * Math.PI / 180;
   }

   function _toDegrees ( radians ) {
      return radians * 180 / Math.PI;
   }

Angular 9中为我工作

  import {GoogleMap, GoogleMapsModule} from "@angular/google-maps";
  @ViewChild('Map') Map: GoogleMap; /* Element Map */

  locations = [
   { lat: 7.423568, lng: 80.462287 },
   { lat: 7.532321, lng: 81.021187 },
   { lat: 6.117010, lng: 80.126269 }
  ];

  constructor() {
   var bounds = new google.maps.LatLngBounds();
    setTimeout(() => {
     for (let u in this.locations) {
      var marker = new google.maps.Marker({
       position: new google.maps.LatLng(this.locations[u].lat, 
       this.locations[u].lng),
      });
      bounds.extend(marker.getPosition());
     }

     this.Map.fitBounds(bounds)
    }, 200)
  }

并根据指示的位置自动将地图居中。

结果:

在此处输入图片说明

我使用上面的方法来设置地图边界,然后,我无需计算缩放级别,而只是计算平均LAT和平均LON并将中心点设置到该位置。我将所有lat值加到latTotal中,将所有lon值加到lontotal中,然后除以标记数。然后,将地图中心点设置为这些平均值。

latCenter = latTotal / markercount; lonCenter = lontotal / markercount;

我遇到了无法更改旧代码的情况,因此添加了此javascript函数以计算中心点和缩放级别:

//input
var tempdata = ["18.9400|72.8200-19.1717|72.9560-28.6139|77.2090"];

function getCenterPosition(tempdata){
	var tempLat = tempdata[0].split("-");
	var latitudearray = [];
	var longitudearray = [];
	var i;
	for(i=0; i<tempLat.length;i++){
		var coordinates = tempLat[i].split("|");
		latitudearray.push(coordinates[0]);
		longitudearray.push(coordinates[1]);
	}
	latitudearray.sort(function (a, b) { return a-b; });
	longitudearray.sort(function (a, b) { return a-b; });
	var latdifferenece = latitudearray[latitudearray.length-1] - latitudearray[0];
	var temp = (latdifferenece / 2).toFixed(4) ;
	var latitudeMid = parseFloat(latitudearray[0]) + parseFloat(temp);
	var longidifferenece = longitudearray[longitudearray.length-1] - longitudearray[0];
	temp = (longidifferenece / 2).toFixed(4) ;
	var longitudeMid = parseFloat(longitudearray[0]) + parseFloat(temp);
	var maxdifference = (latdifferenece > longidifferenece)? latdifferenece : longidifferenece;
	var zoomvalue;	
	if(maxdifference >= 0 && maxdifference <= 0.0037)  //zoom 17
		zoomvalue='17';
	else if(maxdifference > 0.0037 && maxdifference <= 0.0070)  //zoom 16
		zoomvalue='16';
	else if(maxdifference > 0.0070 && maxdifference <= 0.0130)  //zoom 15
		zoomvalue='15';
	else if(maxdifference > 0.0130 && maxdifference <= 0.0290)  //zoom 14
		zoomvalue='14';
	else if(maxdifference > 0.0290 && maxdifference <= 0.0550)  //zoom 13
		zoomvalue='13';
	else if(maxdifference > 0.0550 && maxdifference <= 0.1200)  //zoom 12
		zoomvalue='12';
	else if(maxdifference > 0.1200 && maxdifference <= 0.4640)  //zoom 10
		zoomvalue='10';
	else if(maxdifference > 0.4640 && maxdifference <= 1.8580)  //zoom 8
		zoomvalue='8';
	else if(maxdifference > 1.8580 && maxdifference <= 3.5310)  //zoom 7
		zoomvalue='7';
	else if(maxdifference > 3.5310 && maxdifference <= 7.3367)  //zoom 6
		zoomvalue='6';
	else if(maxdifference > 7.3367 && maxdifference <= 14.222)  //zoom 5
		zoomvalue='5';
	else if(maxdifference > 14.222 && maxdifference <= 28.000)  //zoom 4
		zoomvalue='4';
	else if(maxdifference > 28.000 && maxdifference <= 58.000)  //zoom 3
		zoomvalue='3';
	else
		zoomvalue='1';
	return latitudeMid+'|'+longitudeMid+'|'+zoomvalue;
}

如果有人遇到此线程,这是我的看法:

这有助于防止非数字数据破坏确定lat和的两个最终变量lng

它的工作原理,采取在所有的坐标,把它们解析成单独latlng一个数组的元素,然后确定平均每家的。该平均值应该是中心(在我的测试案例中已证明是正确的。)

var coords = "50.0160001,3.2840073|50.014458,3.2778274|50.0169713,3.2750587|50.0180745,3.276742|50.0204038,3.2733474|50.0217796,3.2781737|50.0293064,3.2712542|50.0319918,3.2580816|50.0243287,3.2582281|50.0281447,3.2451177|50.0307925,3.2443178|50.0278165,3.2343882|50.0326574,3.2289809|50.0288569,3.2237612|50.0260081,3.2230589|50.0269495,3.2210104|50.0212645,3.2133541|50.0165868,3.1977592|50.0150515,3.1977341|50.0147901,3.1965286|50.0171915,3.1961636|50.0130074,3.1845098|50.0113267,3.1729483|50.0177206,3.1705726|50.0210692,3.1670394|50.0182166,3.158297|50.0207314,3.150927|50.0179787,3.1485753|50.0184944,3.1470782|50.0273077,3.149845|50.024227,3.1340514|50.0244172,3.1236235|50.0270676,3.1244474|50.0260853,3.1184879|50.0344525,3.113806";

var filteredtextCoordinatesArray = coords.split('|');    

    centerLatArray = [];
    centerLngArray = [];


    for (i=0 ; i < filteredtextCoordinatesArray.length ; i++) {

      var centerCoords = filteredtextCoordinatesArray[i]; 
      var centerCoordsArray = centerCoords.split(',');

      if (isNaN(Number(centerCoordsArray[0]))) {      
      } else {
        centerLatArray.push(Number(centerCoordsArray[0]));
      }

      if (isNaN(Number(centerCoordsArray[1]))) {
      } else {
        centerLngArray.push(Number(centerCoordsArray[1]));
      }                    

    }

    var centerLatSum = centerLatArray.reduce(function(a, b) { return a + b; });
    var centerLngSum = centerLngArray.reduce(function(a, b) { return a + b; });

    var centerLat = centerLatSum / filteredtextCoordinatesArray.length ; 
    var centerLng = centerLngSum / filteredtextCoordinatesArray.length ;                                    

    console.log(centerLat);
    console.log(centerLng);

    var mapOpt = {      
    zoom:8,
    center: {lat: centerLat, lng: centerLng}      
    };
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