如何更新谷歌地图中多个标记的位置
我正在使用谷歌地图 API 在地图上放置标记。标记的 GPS 坐标存储在 mySQL 数据库中。我已经能够创建标记,但是位置会不断变化,所以我想知道如何更新标记的位置,以便标记能够在地图上移动。到目前为止,这是我的代码:
<!DOCTYPE html>
<html>
<head><title>Map</title>
<style type="text/css">
html { height: 100% }
body { height: 100%; margin: 0px; padding: 0px }
#map_canvas { height: 100% }
</style>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false">
</script>
<script type="text/javascript">
function initialize() {
var latlng = new google.maps.LatLng(36.9947935, -122.0622702);
var myOptions = {
zoom: 15,
center: latlng,
mapTypeId: google.maps.MapTypeId.ROADMAP
};
var map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
var image = new google.maps.MarkerImage('busIcon.png',
// The image size
new google.maps.Size(44, 46),
// The origin
new google.maps.Point(0,0),
// The anchor
new google.maps.Point(22, 23));
var shadow = new google.maps.MarkerImage('busIcon_shadow.png',
new google.maps.Size(58, 46),
new google.maps.Point(0,0),
new google.maps.Point(22, 23)
);
var shape = {
coord: [1, 1, 1, 45, 43, 45, 43 , 1],
type: 'poly'
};
var markers = [
<?php
// Make a MySQL Connection
mysql_connect("**********", "**********", "**********") or die(mysql_error());
mysql_select_db("matsallen") or die(mysql_error());
//Get number of rows
$result = mysql_query
(
'SELECT COUNT(*) FROM busTrack AS count'
)
or die(mysql_error());
$row = mysql_fetch_array( $result );
$length = $row["COUNT(*)"];
for ($count = 1; $count <= $length; $count++) {
//Get data from MySQL database
$result = mysql_query
(
'SELECT * FROM busTrack WHERE busID = '.$count
)
or die(mysql_error());
// store the record of the "busTrack" table into $row
$row = mysql_fetch_array( $result );
// Echo the data into the array 'markers'
$output = '{id: "Bus '.$row[busID].'", lat: '.$row[lat].', lng: '.$row[lon].'}';
if ($count != $length) {
$output = $output.',';
};
echo $output;
};
?>
];
for (index in markers) addMarker(map, markers[index]);
function addMarker(map, data) {
//create the markers
var marker = new google.maps.Marker({
position: new google.maps.LatLng(data.lat, data.lng),
map: map,
title: data.id,
icon: image,
shape: shape,
shadow: shadow
});
//create the info windows
var content = document.createElement("DIV");
var title = document.createElement("DIV");
title.innerHTML = data.id;
content.appendChild(title);
var infowindow = new google.maps.InfoWindow({
content: content
});
// Open the infowindow on marker click
google.maps.event.addListener(marker, "click", function() {
infowindow.open(map, marker);
});
}
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="width:75%; height:75%; margin:10%; allign:center;"></div>
</body>
</html>
I am using google maps api to place markers on a map. The gps coordinates of the markers are stored in a mySQL database. I have been able to create the markers however the locations will constantly changing so I was wondering how I would go about updating the markers' locations so that the markers will move across the map. Here is my code so far:
<!DOCTYPE html>
<html>
<head><title>Map</title>
<style type="text/css">
html { height: 100% }
body { height: 100%; margin: 0px; padding: 0px }
#map_canvas { height: 100% }
</style>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false">
</script>
<script type="text/javascript">
function initialize() {
var latlng = new google.maps.LatLng(36.9947935, -122.0622702);
var myOptions = {
zoom: 15,
center: latlng,
mapTypeId: google.maps.MapTypeId.ROADMAP
};
var map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
var image = new google.maps.MarkerImage('busIcon.png',
// The image size
new google.maps.Size(44, 46),
// The origin
new google.maps.Point(0,0),
// The anchor
new google.maps.Point(22, 23));
var shadow = new google.maps.MarkerImage('busIcon_shadow.png',
new google.maps.Size(58, 46),
new google.maps.Point(0,0),
new google.maps.Point(22, 23)
);
var shape = {
coord: [1, 1, 1, 45, 43, 45, 43 , 1],
type: 'poly'
};
var markers = [
<?php
// Make a MySQL Connection
mysql_connect("**********", "**********", "**********") or die(mysql_error());
mysql_select_db("matsallen") or die(mysql_error());
//Get number of rows
$result = mysql_query
(
'SELECT COUNT(*) FROM busTrack AS count'
)
or die(mysql_error());
$row = mysql_fetch_array( $result );
$length = $row["COUNT(*)"];
for ($count = 1; $count <= $length; $count++) {
//Get data from MySQL database
$result = mysql_query
(
'SELECT * FROM busTrack WHERE busID = '.$count
)
or die(mysql_error());
// store the record of the "busTrack" table into $row
$row = mysql_fetch_array( $result );
// Echo the data into the array 'markers'
$output = '{id: "Bus '.$row[busID].'", lat: '.$row[lat].', lng: '.$row[lon].'}';
if ($count != $length) {
$output = $output.',';
};
echo $output;
};
?>
];
for (index in markers) addMarker(map, markers[index]);
function addMarker(map, data) {
//create the markers
var marker = new google.maps.Marker({
position: new google.maps.LatLng(data.lat, data.lng),
map: map,
title: data.id,
icon: image,
shape: shape,
shadow: shadow
});
//create the info windows
var content = document.createElement("DIV");
var title = document.createElement("DIV");
title.innerHTML = data.id;
content.appendChild(title);
var infowindow = new google.maps.InfoWindow({
content: content
});
// Open the infowindow on marker click
google.maps.event.addListener(marker, "click", function() {
infowindow.open(map, marker);
});
}
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="width:75%; height:75%; margin:10%; allign:center;"></div>
</body>
</html>
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要更改标记的位置,请在该标记上调用 setPosition():
为此,您需要将所有标记保存在数组中。也许这样的事情会起作用:
然后,当标记的位置需要改变时,希望你知道总线ID并且可以调用:
我没有测试上面的代码,所以可能会有小的语法错误或其他东西,但它应该给你这个想法。
To change the location of a marker, call setPosition() on that marker:
To do this, you will need to save all your markers in an array. Perhaps something like this would work:
Then, when the position of a marker needs to change, hopefully you know the bus ID and can just call:
I didn't test the above code so there may be small syntax errors or something, but it should give you the idea.