Javascript Bing映射v8事件传递的pin不正确？

因此，我正在使用bing maps v8编写一个应用程序。不幸的是，它没有z索引支持（尽管有z索引函数…）。所以我尝试使用层来完成类似的事情。问题是，当我在层之间交换pin时，事件开始在具有错误元数据的正确pin位置上触发

下面是我如何创建pin的

function createImagePin(location, index, objArray)
{
    var obj = objArray[index];
    var smallPin = getSmallPin(obj);

    var pin = new Microsoft.Maps.Pushpin(location,
    {
        visible: true,
        icon: smallPin.data,
        anchor: new Microsoft.Maps.Point(smallPin.width / 2, smallPin.height / 2) //Align center of pushpin with location.
    });

    pin.metadata = {};
    pin.metadata.index = index;
    pin.metadata.dataTarget = obj;
    pin.metadata.isSelected = false;

    Microsoft.Maps.Events.addHandler(pin, 'click', function (e) {

        if(e.targetType === 'pushpin')
        {
            SetRowSelection(e.target.metadata.index);
        }
    });

    Microsoft.Maps.Events.addHandler(pin, 'mouseover', function (e)
    {
        if(e.targetType === 'pushpin')
        {
            SelectPin(e.target); 
            ShowInfobox(e.target);
            console.log(e.target.getLocation());
        }
    });

    Microsoft.Maps.Events.addHandler(pin, 'mouseout', function (e)
    {
        if (e.targetType === 'pushpin')
        {
            DeselectPin(e.target);
            HideInfobox();
        }
    });

    return pin;
}

出于某种原因，每次鼠标中有一个pin对象的位置正确，但元数据不正确时，我都会得到它。如果我注释掉取消选择。。。我两个都得到了正确的答案

下面是我如何选择和取消选择引脚。根据给我的例子

目前没有其他代码接触这些引脚。。。我觉得这很奇怪。这几乎就像在内部，它只是复制层中的一些信息，而不是移动附着到对象的所有项目。事实上，如果我只是注释掉add/remove，它就会停止向我传递错误的数据。。。虽然我失去了我的z-索引

function SelectPin(pin)
{
    switch (pin.metadata.dataTarget.ComparableType)
    {
        case "Subject":
            mapLayers.subjectLayer.remove(pin);
            pin.setOptions({ visible: true, icon: largeGreenHouse.data, anchor: new Microsoft.Maps.Point(largeGreenHouse.width / 2, largeGreenHouse.height / 2) });
            break;
        case "Listing Comp":
            mapLayers.compLayer.remove(pin);
            pin.setOptions({ visible: true, icon: largeBlueHouse.data, anchor: new Microsoft.Maps.Point(largeBlueHouse.width / 2, largeBlueHouse.height / 2) });
            break;
        case "Sales Comp":
            mapLayers.compLayer.remove(pin);
            pin.setOptions({ visible: true, icon: largeBlueHouse.data, anchor: new Microsoft.Maps.Point(largeBlueHouse.width / 2, largeBlueHouse.height / 2) });
            break;
        case "Hidden":
            pin.setOptions({ visible: false });
            return;
        default:
            mapLayers.observableLayer.remove(pin);
            pin.setOptions({ visible: true, icon: largeGreyHouse.data, anchor: new Microsoft.Maps.Point(largeGreyHouse.width / 2, largeGreyHouse.height / 2) });
            break;
    }

    mapLayers.selectionLayer.add(pin);
    pin.metadata.isSelected = true;
}

function DeselectPin(pin)
{
    if(!pin.metadata.isSelected)
        return;

    pin.metadata.isSelected = false;
    mapLayers.selectionLayer.remove(pin);  
    switch (pin.metadata.dataTarget.ComparableType)
    {
        case "Subject":
            pin.setOptions({ visible: true, icon: greenHouse.data, anchor: new Microsoft.Maps.Point(greenHouse.width / 2, greenHouse.height / 2) });
            mapLayers.subjectLayer.add(pin);
            break;
        case "Listing Comp":
            pin.setOptions({ visible: true, icon: blueHouse.data, anchor: new Microsoft.Maps.Point(blueHouse.width / 2, blueHouse.height / 2) });
            mapLayers.compLayer.add(pin);
            break;
        case "Sales Comp":
            pin.setOptions({ visible: true, icon: blueHouse.data, anchor: new Microsoft.Maps.Point(blueHouse.width / 2, blueHouse.height / 2) });
            mapLayers.compLayer.add(pin);
            break;
        case "Hidden":
            pin.setOptions({ visible: false });
            return;
        default:
            pin.setOptions({ visible: true, icon: greyHouse.data, anchor: new Microsoft.Maps.Point(greyHouse.width / 2, greyHouse.height / 2) });
            mapLayers.observableLayer.add(pin);
            break;
    }

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嗯，这似乎是数据问题而不是代码问题。我的数据被地理编码为相同的lat/long，用于一些地址不相同的属性。因此，阵列的重新排列会导致相同位置的不同阵列获得“顶部位置”