将php json传递给Google Map API的javascript对象

Question

我建立了一个坐标和其他飓风信息数据库。

<?php
    include '/database_connector.php'; //include database connection information
 $new = array();
 $result=mysqli_query($con, "select * from Spots15 where `Name` = 'Ana' or 'One'");
 while($row=mysqli_fetch_array($result)){
  $lat      = $row['LAT'];
  $long     = $row['LONG'];
  $latlong = $lat.", ". $long;
  array_push($new, $latlong);
}    
  echo json_encode($new);
?>

这是正确的输出。 这是api想要的输出

["31.5, -77.6","31.5, -77.7","31.5, -77.5","31.6, -77.8","31.5, -77.5","31.5, -77.3","31.6, -77.3","31.7, -77.4","31.9, -77.3","32.1, -77.4","32.2, -77.5","32.4, -77.6","32.6, -77.8","32.7, -77.9","32.7, -78.1","32.9, -78.3","32.9, -78.3","33.1, -78.2","33.2, -78.3","33.6, -78.5","33.8, -78.7","34.0, -78.9","34.1, -78.9","34.1, -78.9","34.4, -78.6"]

我想将这些传递给Google Map API，以便在地图上绘制坐标。

var four=new google.maps.LatLng(28.2,-96.0);

Answer 1

您可以在html标记之前（或在另一个PHP页面中，使用include函数）进行数据检索，然后可以使用此行将PHP数组传递给JavaScript：

var arr = <?php echo json_encode($new) ?>;

此时，在for循环中，您可以通过以下方式获得纬度和经度：

var lat = Number(arr[i].split(",")[0].trim());
var lng = Number(arr[i].split(",")[1].trim());

之后，您可以使用变量来绘制所有点。

var four = new google.maps.LatLng(lat, lng);

希望我能正确理解您的问题。

这里有一些代码预览。

var arr = <?php echo json_encode($new) ?>;
for(var i=0; i<arr.length;i++)
{
    var lat = Number(arr[i].split(",")[0].trim());
    var lat = Number(arr[i].split(",")[0].trim());
    var four = new google.maps.LatLng(lat,lng);
    var marker=new google.maps.Marker({
                position:four,
                map:map
            });
}