如何使用php-mysql,js在谷歌地图上显示多个位置

问题描述 投票:-1回答:1

我必须尝试从db获取lat,lng值并像这样在json中编码结果

[{"name":"ahmedabad,GUJ","lat":"72.5226896","lng":"72.5226896"}]

但标记不在地图上显示(位置不在地图上显示)

这是我的代码,我的错误请指导我

<?php

include "db.php";


    /* lat/lng data will be added to this array */
    $locations=array();
    $query =  $db->query("SELECT * FROM ds_petroling_history");
        while( $row = $query->fetch_assoc() ){

            $nama_kabkot = $row['name'];
            $longitude = $row['lng'];                              
            $latitude = $row['lat'];

            /* Each row is added as a new array */
            $locations[]=array( 'name'=>$nama_kabkot, 'lat'=>$longitude, 'lng'=>$longitude );
        }
        /* Convert data to json */
        $markers = json_encode($locations);
?>


<html>
    <head>


    </head>
    <body>

       <div><?php echo $markers ?></div> 


     <script type='text/javascript'>
    <?php
        echo "var markers=$markers;\n";

    ?>

    function initMap() {

        var latlng = new google.maps.LatLng(23.0117523,72.5226665);
        var myOptions = {
            zoom: 10,
            center: latlng,
            mapTypeId: google.maps.MapTypeId.ROADMAP,
            mapTypeControl: false
        };

        var map = new google.maps.Map(document.getElementById("peta"),myOptions);
        var infowindow = new google.maps.InfoWindow(), marker, lat, lng;

        var json=JSON.parse( markers );

        for( var o in json ){

            lat = json[ o ].lat;
            lng=json[ o ].lng;
            name=json[ o ].name;

            marker = new google.maps.Marker({
                position: new google.maps.LatLng(lat,lng),
                name:name,
                map: map
            }); 
            google.maps.event.addListener( marker, 'click', function(e){
                infowindow.setContent( this.name );
                infowindow.open( map, this );
            }.bind( marker ) );
        }
    }
    </script>  
       <div id="peta" style="width: 100%; height: 660px;">
       </div>

    <script async defer src="https://maps.googleapis.com/maps/api/js?key=yourkey&callback=initMap"
  type="text/javascript"></script>
    </body>
</html>

从db成功获取的值并在json中编码但在地图上找不到那些坐标位置

google-maps geolocation google-maps-markers
1个回答
1
投票

你刚刚将标记的值分配给php var而不是javascript var

    /* Convert data to json */
    $markers = json_encode($locations);
    echo "<script>
            var markers[] = " . $markers .";

          </script>";
?>
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