單個航點工作，多個航點不，任何想法？

Question

此代碼是爲我編寫的，用於工作，但現在多個航點不工作。 航點存儲爲| 51.105166，-1.695971 | 51.105166，-1.695971 |在數據庫中。 只有1個航點的地圖工作，但不止一個，他們根本沒有出現。

有沒有人有任何建議？

它可以在這裏的鏈接查看。 http://www.ridersguide.co.uk/2012/Ride_234

var directionDisplay; 
var geocoder; 
var directionsService = new google.maps.DirectionsService(); 
var latlng = new google.maps.LatLng(54.559322587438636, -4.1748046875); 
function load() { 
    geocoder = new google.maps.Geocoder(); 
    directionsDisplay = new google.maps.DirectionsRenderer(); 
    var myOptions = { 
    zoom: 6, 
    mapTypeId: google.maps.MapTypeId.ROADMAP, 
center: latlng, 
    mapTypeControl: true, 
    mapTypeControlOptions: { 
    style: google.maps.MapTypeControlStyle.HORIZONTAL_BAR, 
    position: google.maps.ControlPosition.TOP_RIGHT 
    },  

navigationControl: true, 
navigationControlOptions: { 
style: google.maps.NavigationControlStyle.ZOOM_PAN, 
position: google.maps.ControlPosition.TOP_LEFT} 
    }; 
    var map = new google.maps.Map(document.getElementById('map'), myOptions); 
    directionsDisplay.setMap(map); 
    directionsDisplay.setPanel(document.getElementById("directionsPanel")); 

var directionRendererOptions ={ 
    suppressMarkers: true, 
    polylineOptions:{ 
     strokeColor: "#FF0000", 
     strokeOpacity: 1, 
     strokeWeight: 3 
     } 
    }; 

directionsDisplay.setOptions(directionRendererOptions); 


      var start = '<?php echo $start; ?>'; 
      var end = '<?php echo $end; ?>'; 
      <?php 
      if($via != null){ 
      echo "var points = ["; 
      foreach($via as $point){ 
      if($point != ""){ 
      echo "{location:"; 
      echo " '".$point."'"; 
      echo "}"; 
      } 
      } 
      echo "];\n"; 
      } 
      ?> 

      var request = { 
       origin:start, 
       waypoints: points, 
       destination:end, 
       travelMode: google.maps.DirectionsTravelMode.DRIVING 
      }; 
      directionsService.route(request, function(response, status) { 
       if (status == google.maps.DirectionsStatus.OK) { 
        directionsDisplay.setDirections(response); 

       } 

      }); 

     geocoder.geocode({ 'address': start}, function(results, status) { 
     var routeStart = new google.maps.Marker({ 
     map: map, 
     position: results[0].geometry.location, 
     icon: './images/motorcycling.png', 
     shadow: './images/motorcycling.shadow.png' 
    }); 

}); 
    geocoder.geocode({ 'address': end}, function(results, status) { 
     var routeEnd = new google.maps.Marker({ 
     map: map, 
     position: results[0].geometry.location, 
     icon: './images/motorcyclingend.png', 
     shadow: './images/motorcycling.shadow.png' 
    }); 

}); 
<?php` 

Answer 1

這似乎有一個逗號陣列中丟失。

var points = [{location: '51.105166,-1.695971'}{location: '51.105166,-1.695971'}]; 

應該做的

var points = [{location: '51.105166,-1.695971'},{location: '51.105166,-1.695971'}]; 

一種方式，是完全構建字符串在PHP中，而不是將它輸出零零碎碎，然後刪除不需要的初始逗號。我已經編輯這個代碼以下收到的意見，所以只在被構造[]和逗號除去之間的位置：

<?php 
      if($via != null){ 
      $pointstring=''; 
      foreach($via as $point){ 
      if($point != ""){ 
      $pointstring.= ",{location:"; 
      $pointstring.= " '".$point."'"; 
      $pointstring.= "}"; 
      } 
      } 
      echo "var points = [".substr($pointstring,1)."];\n"; 
      } 
      ?> 

（代碼未測試）

有輸出該位置的其它方式就像你曾經有過的那樣，但是要求在它們之間有逗號，而不是在最後，這會比簡單地移除一個你不需要的更復雜。