Question

我正在尝试显示从中调用的地址我的sql数据库，我无法显示我一直收到此错误的地址＆＃34;地理编码由于以下原因而无法成功Zero_Results＆＃34;

我试图在一个变量$地址中添加完整地址，而不是使用三个变量$ address，$ county，$ county但仍然没有工作，我无法理解。

我想将地址显示为标记，如图所示。

每当我在我的数据库中添加新地址时，我都会在地图中添加新标记。

enter image description here

任何帮助，

提前致谢。

以下是我正在使用的代码：

$result = mysqli_query($con,"SELECT * FROM `test`") or die ("Error:      
 ".mysqli_error($con));
     while ($row = mysqli_fetch_array($result))
     {
            $address = $row['address'];
            $county = $row['County'];
            $country = $row['Country'];
     }
        mysqli_close($con); 

        <!DOCTYPE html>
         <html>
          <head>

       <script type="text/javascript"
           src="https://maps.googleapis.com/maps/api/js?                       
           key=mykey&sensor=false">

        </script>


         <script type="text/javascript">
          var geocoder;
          var map;
          function initialize() {
            geocoder = new google.maps.Geocoder();
            var latlng = new google.maps.LatLng(          
            53.41291,-8.243889999999965);
            var address = '<?php echo $address.', '.$country.', '.$county; ?                 
          >';

            var myOptions = {
              zoom: 6,
              center: latlng,
              mapTypeId: google.maps.MapTypeId.ROADMAP
            }
            map = new google.maps.Map(document.getElementById("map_canvas"),                  
             myOptions);
            geocoder.geocode( { 'address': address}, function(results,    
              status) {
              if (status == google.maps.GeocoderStatus.OK) {
                map.setCenter(results[0].geometry.location);
                var marker = new google.maps.Marker({
                    map: map, 
                    position: results[0].geometry.location
                });
              } else {
                alert("Geocode was not successful for the following reason:                    
               " + status);
              }
            });
          }
        </script>
        </head>
        <body onload="initialize()">
          <div id="map_canvas" style="width:500px; height:500px"></div>

        </body>
        </html>

Answer 1

我的问题解决了，我的解决方案是：

            <script type="text/javascript">
            var geocoder;
            var map;
            function initialize(address) {
            geocoder = new google.maps.Geocoder();
            var latlng = new google.maps.LatLng(   
              53.41291,-8.243889999999965);


            var myOptions = {
              zoom: 6,
              center: latlng,
              mapTypeId: google.maps.MapTypeId.ROADMAP
            }
            map = new google.maps.Map(document.getElementById("map_canvas"),            
            myOptions);
            <?php
            require('connection.php');

              $result = mysqli_query($con,"SELECT * FROM `test`") or die  
                ("Error: ".mysqli_error($con));
              $count = 0;


            ?>
            <?php  //Starts while loop so all addresses for the given 
                // information will be populated.
             $addresscounting=0;
               while($row = mysqli_fetch_assoc($result)) //instantiates 
                 // array
           { ?>   


             var address = "<?php echo $address ?>";


            geocoder.geocode( { 'address': address}, function(results,                 
             status) {
              if (status == google.maps.GeocoderStatus.OK) {
                map.setCenter(results[0].geometry.location);
                var marker = new google.maps.Marker({
                    map: map, 
                    position: results[0].geometry.location
                });
              } else {
                alert("Geocode was not successful for the following reason:             
             " + status);
              }
            });

         <?php
               $count++;

              } //ends while
          ?>
          }

          </script

由于以下原因，地理编码不成功零结果 - PHP变量

1 个答案: