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问题：如何更改我的代码，以便我可以使用普通的人类地址，并在地图上将它们“标记”。

好的 - 我已经为此工作了一段时间，但我似乎无法做到这一点。我有以下代码在地图上显示自定义标记 - 100% 正确工作：

function LoadMap() {
    var locations = new Array(
        [34.01843,-118.491046], [34.006673,-118.486562], [34.009714,-118.480296]
    );

    var markers = new Array();

    var mapOptions = {
        center: new google.maps.LatLng(34.019000, -118.455458),
        zoom: 14,
        mapTypeId: google.maps.MapTypeId.ROADMAP,
        scrollwheel: false
    };

    var map = new google.maps.Map(document.getElementById('map'), mapOptions);

    $.each(locations, function(index, location) {
        var marker = new google.maps.Marker({
            position: new google.maps.LatLng(location[0], location[1]),
            map: map
        });

        var myOptions = {
            content: '<div class="infobox"><div class="image"><img src="http://html.realia.byaviators.com/assets/img/tmp/property-tiny-1.png" alt=""></div><div class="title"><a href="detail.html">1041 Fife Ave</a></div><div class="area"><span class="key">Area:</span><span class="value">200m<sup>2</sup></span></div><div class="price">€450 000.00</div><div class="link"><a href="detail.html">View more</a></div></div>',
            disableAutoPan: false,
            maxWidth: 0,
            pixelOffset: new google.maps.Size(-146, -190),
            zIndex: null,
            closeBoxURL: "",
            infoBoxClearance: new google.maps.Size(1, 1),
            position: new google.maps.LatLng(location[0], location[1]),
            isHidden: false,
            pane: "floatPane",
            enableEventPropagation: false
        };
        marker.infobox = new InfoBox(myOptions);
        marker.infobox.isOpen = false;

        var myOptions = {
            draggable: true,
            content: '<div class="marker"><div class="marker-inner"></div></div>',
            disableAutoPan: true,
            pixelOffset: new google.maps.Size(-21, -58),
            position: new google.maps.LatLng(location[0], location[1]),
            closeBoxURL: "",
            isHidden: false,
            enableEventPropagation: true
        };
        marker.marker = new InfoBox(myOptions);
        marker.marker.open(map, marker);
        markers.push(marker);

        google.maps.event.addListener(marker, "click", function (e) {
            var curMarker = this;

            $.each(markers, function (index, marker) {
                // if marker is not the clicked marker, close the marker
                if (marker !== curMarker) {
                    marker.infobox.close();
                    marker.infobox.isOpen = false;
                }
            });

            if(curMarker.infobox.isOpen === false) {
                curMarker.infobox.open(map, this);
                curMarker.infobox.isOpen = true;
                map.panTo(curMarker.getPosition());
            } else {
                curMarker.infobox.close();
                curMarker.infobox.isOpen = false;
            }

        });
    });
}

我想要做的是将“var 位置”从 lat/long 更改为 json 'human' 地址 - 从此：

var locations = new Array(
        [34.01843,-118.491046], [34.006673,-118.486562], [34.009714,-118.480296]
    );

像这样：

var locations = JSON.parse('[{"address":"New Street, Salisbury, UK","content":"hello world from salisbury","status":"live"},{"address":"86000 Poitiers, France","content":"hello world from poitiers"},{"address":"66000 Perpignam, France","content":"hello world from perpignam"}]');

所以很明显我需要在使用它们之前以某种方式对这些地址进行地理编码google.maps.Marker position- 但我似乎无法弄清楚如何。

我尝试将标记代码更改为：

    geocoder.geocode( {'address': '1 George St Sydney NSW 2000 Australia' },
              function(results, status) {

                if (status == google.maps.GeocoderStatus.OK) {

                    marker = new google.maps.Marker({
                        position: new google.maps.LatLng(results[0].geometry.location.lat(), results[0].geometry.location.lng()),
                        map: map,
                        icon: 'transparent.png'
                    });
                } 
                else {
                  document.getElementById("text_status").value = status;
                }
              }
            );

但是没有渲染标记:(

任何帮助/建议将不胜感激。

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0 回答 0