将标记从谷歌地图v3放在所有其他人面前 [英] Put marker from Google Maps v3 in front of all others

问题描述

有人可以帮助我把当前位置放在所有其他位置？我已经阅读了MAX_ZINDEX和setZindex（），但我无法理解它。这是我的代码：

  var latlng = new google.maps.LatLng（lat，lng）; 
 var image =; 
 var currentplace =Ohla Hotel; 
 var zInd =; 
 if（name == currentplace）{
 image =../img/hotel_icon.png; 
} 
 else {
 image =../img/home_icon.png; 
} 
 // alert（）; 
 var maxZindex = google.maps.Marker.MAX_ZINDEX; 
 if（name == currentplace）{
 zInd =（maxZindex + 1）; 
} 
 var locationDescription = this.locationDescription; 
 var marker = new google.maps.Marker（{
 map：map，
 position：latlng，
 title：'pk：'+ name，
 MAX_ZINDEX： zInd，
 icon：image 
}）; 
 bounds.extend（latlng）; 
 setMarker（map，marker，name，locationDescription）; 
  

解决方案

MAX_ZINDEX

相关的MarkerOption是 zIndex ：

p>

  var marker = new google.maps.Marker（{
 map：map，
 position：latlng，
 title：'pk：'+ name，
 zIndex：zInd，
 icon：image 
}）; 
  

这会将标记的 zIndex 属性设置为你已经计算出的价值，比API的最大值多一个。

Can somebody help me put the current location in front of all others? I have read about MAX_ZINDEX and setZindex() but I cannot understand it. This is my code:

var latlng = new google.maps.LatLng(lat,lng);
var image = "";
var currentplace = "Ohla Hotel";
var zInd = "";
if( name == currentplace ) {
    image = "../img/hotel_icon.png";
}
else {
    image = "../img/home_icon.png";
}
//alert();
var maxZindex = google.maps.Marker.MAX_ZINDEX;
if( name == currentplace ) {
    zInd = (maxZindex + 1);
}
var locationDescription = this.locationDescription;
var marker = new google.maps.Marker({
    map: map,
    position: latlng,
    title:'pk:'+name,
    MAX_ZINDEX: zInd,
    icon: image
});
bounds.extend(latlng);
setMarker(map, marker, name, locationDescription);

解决方案

MAX_ZINDEXis a constant, not an attribute of a marker.

The relevant MarkerOption is zIndex:

var marker = new google.maps.Marker({
    map: map,
    position: latlng,
    title:'pk:'+name,
    zIndex: zInd,
    icon: image
});

This sets the marker's zIndex attribute to the value you have calculated, which is one more than the API's maximum value.

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