我已經到了從通過PHP創建的標記集傳遞一個值的點,但我無法弄清楚如何創建或實現IF條件來顯示標記與「類型」值相關的圖像。基於PHP值的Google Maps API V3中的不同標記
代碼:
<script type="text/javascript">
var iconStar = new google.maps.MarkerImage("googleMarkers/star.png",
new google.maps.Size(32, 28),
new google.maps.Point(0, 0),
new google.maps.Point(16, 32));
var iconBlue = new google.maps.MarkerImage("images/mm_20_blue.png",
new google.maps.Size(12, 20),
new google.maps.Point(0,0),
new google.maps.Point(6, 20));
var iconRed = new google.maps.MarkerImage("images/mm_20_red.png",
new google.maps.Size(12, 20),
new google.maps.Point(6, 20),
new google.maps.Point(5, 1));
var iconYellow = new google.maps.MarkerImage("images/mm_20_yellow.png",
new google.maps.Size(12, 20),
new google.maps.Point(6, 20),
new google.maps.Point(5, 1));
iconType = [] = iconStar;
iconType["0"] = iconStar;
iconType["1"] = iconBlue;
iconType["2"] = iconRed;
iconType["3"] = iconYellow;
var center = null;
var map = null;
var currentPopup;
var bounds = new google.maps.LatLngBounds();
function addMarker(lat, lng, info, type) {
var pt = new google.maps.LatLng(lat, lng);
bounds.extend(pt);
var marker = new google.maps.Marker({
position: pt,
icon: iconType,
map: map
});
var popup = new google.maps.InfoWindow({
content: info,
maxWidth: 300
});
google.maps.event.addListener(marker, "click", function() {
if (currentPopup != null) {
currentPopup.close();
currentPopup = null;
}
popup.open(map, marker);
currentPopup = popup;
});
google.maps.event.addListener(popup, "closeclick", function() {
// panTo puts you back to the original center - not good for zoomed in nav
// map.panTo(center);
currentPopup = null;
});
}
function initMap() {
map = new google.maps.Map(document.getElementById("map"), {
center: new google.maps.LatLng(0, 0),
zoom: 13,
mapTypeId: google.maps.MapTypeId.ROADMAP,
mapTypeControl: false,
mapTypeControlOptions: {
style: google.maps.MapTypeControlStyle.HORIZONTAL_BAR
},
navigationControl: true,
navigationControlOptions: {
style: google.maps.NavigationControlStyle.DEFAULT
}
});
<?php
do {
$name=$row_rsCity['DEALER'];
$lat=$row_rsCity['lat'];
$lon=$row_rsCity['lng'];
$desc=$row_rsCity['ADDRESS'];
$city=$row_rsCity['CITY'];
$state=$row_rsCity['STATE'];
$phone=$row_rsCity['PHONENUMBER'];
$type=$row_rsCity['DEALER_TYPE'];
echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/>$city , $state<br />Phone: $phone',$type);\n");
} while ($row_rsCity = mysql_fetch_assoc($rsCity));
?>
center = bounds.getCenter();
map.fitBounds(bounds);
}
我很接近,但我不能找到類似的例子網上,這樣尋找解決這一問題的幫助不大。
謝謝!
圖標:iconType,變更爲圖標:iconType [型],只返回iconStar或零值,然後iconBlue但未進一步。 更改爲圖標:iconType [「型」], 返回沒有自定義標記只有在默認谷歌的紅色標記 這似乎是正確的,如果沒有包裝在引號作爲Booleen操作開啓或關閉,但傳遞的價值觀的問題當用引號包裝失敗時通過或理解值.. – Burndog 2012-02-02 22:02:06