google-maps - 谷歌地图具有完全相同位置的多个标记不起作用

Question

我想检查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：

<!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>

我厌倦了在这里找到这个工作示例http://www.geocodezip.com/SO_OverQueryLimitB.html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，最终结果将显示一个包含合并文本的信息窗口，如此处http://maps.caseypthomas.org/ex/MarkerClustererPlus/exCoincidentMarkers_SharedInfowindow_wGeocoding.html查看它显示数字 4 但仅显示 3 个标记，这是因为右侧的一个与后面的另一个合并，当您单击它时，它会显示两者的文本。只有我想使用 geocodezip 的示例并在此基础上进行工作，因为我已经有了绳索，不需要谷歌为我获取它们。如果您注意到其他问题，感谢您阅读这个长问题。

如果你能帮我找到解决方案，谢谢 1Mill X。

再次感谢！！！

score 22 · Accepted Answer

你需要：

首先创建标记聚类器。

将标记添加到 MarkerClusterer （并格式化您的代码，使其更易于阅读......）。

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;
}

工作示例

google-maps - 谷歌地图具有完全相同位置的多个标记不起作用

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