我正在使用关于 Google Maps JS API v3 - Simple Multiple Marker Example 问题的信息
我会举报的:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8" />
<title>Google Maps Multiple Markers</title>
<script src="http://maps.google.com/maps/api/js?sensor=false"
type="text/javascript"></script>
</head>
<body>
<div id="map" style="width: 500px; height: 400px;"></div>
<script type="text/javascript">
var locations = [
['Bondi Beach', -33.890542, 151.274856, 4],
['Coogee Beach', -33.923036, 151.259052, 5],
['Cronulla Beach', -34.028249, 151.157507, 3],
['Manly Beach', -33.80010128657071, 151.28747820854187, 2],
['Maroubra Beach', -33.950198, 151.259302, 1]
];
var map = new google.maps.Map(document.getElementById('map'), {
zoom: 10,
center: new google.maps.LatLng(-33.92, 151.25),
mapTypeId: google.maps.MapTypeId.ROADMAP
});
var infowindow = new google.maps.InfoWindow();
var marker, i;
for (i = 0; i < locations.length; i++) {
marker = new google.maps.Marker({
position: new google.maps.LatLng(locations[i][1], locations[i][2]),
map: map
});
google.maps.event.addListener(marker, 'click', (function(marker, i) {
return function() {
infowindow.setContent(locations[i][0]);
infowindow.open(map, marker);
}
})(marker, i));
}
</script>
</body>
</html>
它可以工作,但现在我想改进数据库中的位置。
var locations = [
<?php
while($row=mysql_fetch_array($resultqr)){
echo ("
[".$row[nome].", ".$row[latitudine].", ".$row[longitudine].",],
");
}
?>
];
这段代码在测试页面上有效,但在脚本中的可变位置无效。
数据库连接在脚本之外的其他 php 部分。
你能帮帮我吗?
最佳答案
echo json_encode(array_values($row));
或者,更有保证:
$arr = array($row['nome'], $row['latitudine'], $row['longitudine]');
echo json_encode($arr);
您手工制作的 js 数组中的问题是字符串周围缺少引号。
考虑您的完整 PHP 代码:
<?php
$locations = array();
while($row=mysql_fetch_array($resultqr)){
$locations[] = array($row['nome'], $row['latitudine'], $row['longitudine']);
}
echo 'var locations = ' . json_encode($locations) . ';';
?>
关于javascript - Google Maps JS API v3 - 简单多标记 V2,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20009108/