2017-09-15 234 views
-1

所以我試圖從我的mysqli數據庫顯示來自多個GPS座標的GPS蹤跡/圖。我的php文件顯示地圖很好,但沒有顯示任何標記。有什麼建議麼?另外,因爲我已經問過一個問題,有沒有辦法顯示一行而不是多個標記?PHP在地圖上不顯示標記

 <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

請提供[MCVE]演示這個問題。我們沒有你的數據庫,但發送給瀏覽器的HTML/Javascript/CSS(一個最小樣本)可以工作。如果我從發佈的代碼[[mcve]] [它工作(小提琴)](http://jsfiddle.net/geocodezip/0e8kt98y/) – geocodezip

+0

@geocodezip似乎有什麼問題,我的while循環,它不是射擊,我知道那裏有行,我沒有'$ rowcount = mysqli_num_rows($ result); echo「$ rowcount」;'並返回27行。 – user2101081

回答

0

好了,所以我在while循環,有輕微的錯誤,這是我的新while循環,它的工作..

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