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我的问题是，当我第一次加载页面时它没有出错，但是当我切换回此地图或从另一个选项卡切换回来时，它给了我以下错误消息：无效参数。行：26 字符：56 代码：0 URI：http ://maps.gstatic.com/intl/en_us/mapfiles/api-3/9/9/main.js

请帮助我的代码在这里... :(

 $start1=$makers['lat'];
$start2=$makers['lng'];
$name1=$makers['name'];
$end1=$makere['lat'];
$end2=$makere['lng'];
$name2=$makere['name'];


<script type="text/javascript">
    //<![CDATA[
$(document).ready(function(){ 

var cenLatlng = new google.maps.LatLng(<?php echo $start1;?>, <?php echo $start2;?>);
    var myOptions = {
                       zoom: 15,
                       center: cenLatlng,
                       mapTypeId: google.maps.MapTypeId.ROADMAP
                    }

    var map = new google.maps.Map(document.getElementById("map_det-<?php echo $trip_id;?>"),myOptions);

    var myLatlng = new google.maps.LatLng(<?php echo $start1;?>, <?php echo $start2;?>);
    var marker = new google.maps.Marker({
                                          position: myLatlng,
                                          map: map,
                                          title:"<?php echo $name1;?>"
                                       });

    var contentString = '<?php echo $name1;?>';

    var infowindow = new google.maps.InfoWindow({
                                                  content: contentString,
                                                  width:100,
                                                  height:50
                                               });



    // Start of newly added code block
    google.maps.event.addListener(marker, 'click', function() {
        <?
     $sql2=mysql_fetch_array(mysql_query("select pickpoint from trip_tbl where trip_id='$trip_id'"));
     $pickpoint=$sql2['pickpoint'];
$makers1=mysql_fetch_array(mysql_query("select * from markers where id='$pickpoint'"));
$name1=$makers1['name'];
    ?>
    alert("<?=$name1;?>");

});
});

 //]]>

  </script><div id="map_det-<?php echo $trip_id;?>" style="width:300px; height:200px; border:solid; border-color:#666; "></div>
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