Question

我有这个代码从数据库中获取细节并将它们放在php数组中，稍后将其传递给javascript以生成Google地图：

<?php
$query='select name,longitude,latitude from maps';
$result=mysql_query($query) or die();
$num_rows=mysql_num_rows($result);
if($num_rows > 0){
for($i=0;$i<$num_rows;$i++){
$row=mysql_fetch_row($result);

$location[]= $row[0].', '.$row[2].', '.$row[1].','.($i+1);


}
//this creates the following array
array(4) { 
[0]=> string(44) "Travel agent Marko, 42.5624525, 27.5263493,1" 
[1]=> string(44) "Travel agent Marko, 42.5632957, 27.5267364,2" 
[2]=> string(54) "Travel agent 'Planet Travel', 42.5609811, 27.5263934,3" 
[3]=> string(56) "Travel agent 'Planet Travel' 2, 42.5642942, 27.5266121,4" 
}

?>
<script type="text/javascript">
var loc = <?php echo json_encode($location); ?>;
var locations = [
[loc]
];


var map = new google.maps.Map(document.getElementById('googlemap'), {
  zoom: 10,
  center: new google.maps.LatLng(42.5624525, 27.5263493),
  mapTypeId: google.maps.MapTypeId.ROADMAP
});

var infowindow = new google.maps.InfoWindow();

var marker, i;
alert(locations.length);

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>

<?php
}
?>

我的问题是为什么标记不会出现在地图上？没有返回错误，但也没有标记。我是否必须在php循环中移动javascript代码？如果我提醒locations.length我得到1作为locations.length。为什么js将位置数组计为1个位置？提前谢谢！

Answer 1

改变这个：

UITextField

到此：

var loc = <?php echo json_encode($location); ?>;
var locations = [
    [loc]
];

循环应该是：

var locations = <?php echo json_encode($location); ?>;

谷歌地图 - 来自数据库经度和格度的多个标记未在地图上显示

1 个答案: