Question

我想查看是否有任何现有标记与新标记的latlng匹配，如果是，则合并信息窗口/工具提示文本。

这就是我试图做的事情：

<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8"/>
<title></title>
<script type="text/javascript"  src="http://maps.google.com/maps/api/js?sensor=false"></script>
<script type="text/javascript" src="http://google-maps-utility-library-v3.googlecode.com/svn/trunk/markerclustererplus/src/markerclusterer.js"></script>


<script type="text/javascript"> 
var map;
var mc;//marker clusterer
var mcOptions = {gridSize: 10, maxZoom: 8};
var infowindow = new google.maps.InfoWindow();//global infowindow
var geocoder = new google.maps.Geocoder(); //geocoder
var address = new Array("42.3334,-89.1572",
"39.2058,-76.7531",
"39.7751,-86.1322",
"40.4894,-78.3499",
"42.0203,-87.9059",
"36.2673,-86.2912",
"33.6115,-84.3745",
"44.9793,-93.273",
"40.1461,-76.0738",
"32.2911,-90.1927",
"32.9315,-96.6158",
"36.0553,-79.8317",
"41.8397,-88.0887",
"47.8029,-103.267",
"34.106,-83.589",
"41.5907,-87.3199",
"43.0905,-74.3554",
"40.3438,-74.4289",
"40.8651,-96.8231",
"40.8651,-96.8231",
"41.759,-88.1524",
"38.2512,-86.8675",
"41.8119,-87.6873",
"41.3651,-89.0866",
"25.7791,-80.1978",
"41.6404,-88.0696",
"41.7684,-88.1366",
"39.7299,-86.4234",
"41.5234,-81.5996",
"41.6233,-88.0225",
"41.0171,-80.8029",
"40.2899,-82.9811",
"41.8119,-87.6873",
"32.3445,-99.8021",
"41.8119,-87.6873",
"29.8131,-95.3098",
"35.1693,-89.9904",
"33.6115,-84.3745",
"47.7374,-103.298",
"46.3502,-94.1",
"41.9907,-88.4298",
"35.3716,-80.5621",
"38.189,-85.6768",
"41.8119,-87.6873",
"32.7714,-97.2915");
var content = new Array("UnitNo1",
"UnitNo2",
"UnitNo3",
"UnitNo4",
"UnitNo5",
"UnitNo6",
"UnitNo7",
"UnitNo8",
"UnitNo9",
"UnitNo10",
"UnitNo11",
"UnitNo12",
"UnitNo13",
"UnitNo14",
"UnitNo15",
"UnitNo16",
"UnitNo17",
"UnitNo18",
"UnitNo19",
"UnitNo20",
"UnitNo21",
"UnitNo22",
"UnitNo23",
"UnitNo24",
"UnitNo25",
"UnitNo26",
"UnitNo27",
"UnitNo28",
"UnitNo29",
"UnitNo30",
"UnitNo31",
"UnitNo32",
"UnitNo33",
"UnitNo34",
"UnitNo35",
"UnitNo36",
"UnitNo37",
"UnitNo38",
"UnitNo39",
"UnitNo40",
"UnitNo41",
"UnitNo42",
"UnitNo43",
"UnitNo44",
"UnitNo45");

//min and max limits for multiplier, for random numbers //keep the range pretty small, so markers are kept close by
var min = .999999;
var max = 1.000001;

function createMarker(latlng,text) {

var marker = new google.maps.Marker({
position: latlng,
                map: map
});

                ///get array of markers currently in cluster
                var allMarkers = mc.getMarkers();

                //check to see if any of the existing markers match the latlng of the new marker
                if (allMarkers.length != 0) {
                for (i=0; i < allMarkers.length; i++) {
                var existingMarker = allMarkers[i];
                var pos = existingMarker.getPosition();

                if (latlng.equals(pos)) {
                text = text + " & " + content[i];
                }                   
                }
                }
// WHERE TO ADD: mc.addMarker(marker); //??

google.maps.event.addListener(marker, 'click', function() {
infowindow.close();
infowindow.setContent(text);
infowindow.open(map,marker);
});
return marker;
}

function initialize(){
var options = { 
zoom: 4, 
center: new google.maps.LatLng(39.8282,-98.5795), 
mapTypeId: google.maps.MapTypeId.ROADMAP 
};
map = new google.maps.Map(document.getElementById('map'), options); 

var gmarkers = [];
for (i=0; i<address.length; i++) {
var ptStr = address[i];
var coords = ptStr.split(",");
var latlng = new google.maps.LatLng(parseFloat(coords[0]),parseFloat(coords[1]))
gmarkers.push(createMarker(latlng,content[i]));
}

//marker cluster
mc = new MarkerClusterer(map, gmarkers, mcOptions);
for (i=0; i<address.length; i++) { 
geocodeAddress(address[i],i);

}
}
</script> 
<style>
html, body, #map {
  height: 100%;
  width: 100%;
  margin: 0;
  padding: 0;
  }
</style>
</head>
<body onload="initialize();">
<div id="map"></div>
</body>
</html>

我厌倦了在http://www.geocodezip.com/SO_OverQueryLimitB.html找到这个工作示例，并将以下代码添加到其中：

        ///get array of markers currently in cluster
        var allMarkers = mc.getMarkers();

        //check to see if any of the existing markers match the latlng of the new marker
        if (allMarkers.length != 0) {
        for (i=0; i < allMarkers.length; i++) {
        var existingMarker = allMarkers[i];
        var pos = existingMarker.getPosition();

        if (latlng.equals(pos)) {
        text = text + " & " + content[i];
        }                   
        }
        }

因此，当您单击具有相同latlng的标记时的最终结果将显示一个信息窗口，其中包含合并文本，如此处找到的http://maps.caseypthomas.org/ex/MarkerClustererPlus/exCoincidentMarkers_SharedInfowindow_wGeocoding.html看到它显示数字4但仅显示3个标记这是因为右侧的那个与其后的另一个合并，当您点击它时，它会显示两者的文本。只有我想使用geocodezip的例子并在其上工作，因为我已经有了电线，并且不需要谷歌去为我取得它们。感谢您阅读此LONG问题，如果注意到其他问题......

感谢1Mill X，如果你能帮我找到解决方案。

再次感谢!!!

Answer 1

你需要：

首先创建标记聚类器。

将标记添加到MarkerClusterer（并格式化代码，以便更容易阅读......）。

function createMarker(latlng,text) {
  var marker = new google.maps.Marker({
    position: latlng,
    map: map
  });

  ///get array of markers currently in cluster
  var allMarkers = mc.getMarkers();

  //check to see if any of the existing markers match the latlng of the new marker
  if (allMarkers.length != 0) {
    for (i=0; i < allMarkers.length; i++) {
      var existingMarker = allMarkers[i];
      var pos = existingMarker.getPosition();

      if (latlng.equals(pos)) {
        text = text + " & " + content[i];
      }                   
    }
  }

  google.maps.event.addListener(marker, 'click', function() {
    infowindow.close();
    infowindow.setContent(text);
    infowindow.open(map,marker);
  });
  mc.addMarker(marker);
  return marker;
}

working example

Answer 2

相同的例子我修改并在文本而不是信息窗口中显示数据，如果它是一个它将显示在一个div中，如果它超过一个然后它将显示在旋转木马...请检查示例...

Clustered Maps Multiple markers with the exact same location

Example

谷歌地图具有完全相同位置的多个标记不起作用

2 个答案: