2017-09-15 133 views
-1

所以我试图从我的mysqli数据库显示来自多个GPS坐标的GPS踪迹/图。我的php文件显示地图很好,但没有显示任何标记。有什么建议么?另外,因为我已经问过一个问题,有没有办法显示一行而不是多个标记?PHP在地图上不显示标记

 <html> 
<head> 
<meta http-equiv="content-type" content="text/html; charset=utf-8"/> 
<style type="text/css"> 
    body { font: normal 10pt Helvetica, Arial; } 
#map { width: 600px; height: 600px; border: 0px; padding: 0px; } 
</style> 

<script src="http://maps.google.com/maps/api/js?key=myapikey&v=3&sensor=false" type="text/javascript"></script> 
<script type="text/javascript"> 

var icon = new google.maps.MarkerImage("http://maps.google.com/mapfiles/ms/micons/blue.png", 
new google.maps.Size(32, 32), new google.maps.Point(0, 0), 
new google.maps.Point(16, 32)); 
var center = null; 
var map = null; 
var currentPopup; 
var bounds = new google.maps.LatLngBounds(); 

function addMarker(lat, lng, info) 
{ 
    var pt = new google.maps.LatLng(lat, lng); 
    bounds.extend(pt); 

    var marker = new google.maps.Marker({ 
    position: pt, 
    icon: icon, 
    map: map 
    }); 
} 

function initMap() 
{ 
    map = new google.maps.Map(document.getElementById("map"), 
    { 
     center: new google.maps.LatLng(0, 0), 
     zoom: 14, 
     mapTypeId: google.maps.MapTypeId.ROADMAP, 
     mapTypeControl: false, 
     mapTypeControlOptions: { 
     style: google.maps.MapTypeControlStyle.HORIZONTAL_BAR 
     }, 
     navigationControl: true, 
     navigationControlOptions: { 
     style: google.maps.NavigationControlStyle.SMALL 
     } 
    }); 

<?php 
$sql = "SELECT * FROM gps_points WHERE email = '$email' AND gps_id = '$route_id'"; 
$result = $link->query($sql); 

while ($row = mysql_fetch_array($result)) 
{ 
    $lat=$row['e_lat']; 
    $lon=$row['e_long']; 
    echo ("addMarker($lat, $lon);\n"); 
} 
?> 
center = bounds.getCenter(); 
map.fitBounds(bounds); 
} 
</script> 
</head> 
    <body onload="initMap()" style="margin:0px; border:0px; padding:0px;"> 
    <div id="map"></div> 
</html> 
+0

请提供[MCVE]演示这个问题。我们没有你的数据库,但发送给浏览器的HTML/Javascript/CSS(一个最小样本)可以工作。如果我从发布的代码[[mcve]] [它工作(小提琴)](http://jsfiddle.net/geocodezip/0e8kt98y/) – geocodezip

+0

@geocodezip似乎有什么问题,我的while循环,它不是射击,我知道那里有行,我没有'$ rowcount = mysqli_num_rows($ result); echo“$ rowcount”;'并返回27行。 – user2101081

回答

0

好了,所以我在while循环,有轻微的错误,这是我的新while循环,它的工作..

while($row = $result->fetch_assoc()) 
{ 
    $lat=$row['e_lat']; 
    $lon=$row['e_long']; 
    echo ("addMarker($lat, $lon);\n"); 
}