自动查询MySQL并使用AJAX输出结果

Question

我在Google地图上自动显示我的gps坐标时遇到问题。当我实现地图是静态的时候一切正常，但现在我动态地尝试它。这意味着当新的GPS坐标将被保存到数据库时，地图应该自动刷新。

我在一个.php文件中实现了一切。为了显示与COORDS的地图，用户必须选择一个条目，链接到一个条目是这样的 - >http://trackmyrun1.at/index.php?page=show_runs&data1=get&data2=1

<?php 
    $result = mysql_query("SELECT * FROM fifa_gps WHERE run_id=".$_GET['data2'].""); 

    while ($line = mysql_fetch_array($result)) { 
    $cords[] = "new google.maps.LatLng(" . $line['lat'] . ", " . $line['longt'] . "),"; 
    } 

    $cord_start = explode("(", $cords[0]); 
    $cord_pos_start = explode(")", $cord_start[1]); 

    $cord_end = explode("(", $cords[count($cords)-1]); 
    $cord_pos_end = explode(")", $cord_end[1]); 

    echo $sys->overall_distance($_GET['data2']); 
    ?>  

    <div id="map-canvas" style="height:400px"/> 
    <script type="text/javascript"> 
    var seconds = 1; 
    var divid = "map-canvas"; 
    var url = "http://localhost/alt/FIFA-Europaliga/index.php?page=show_runs&data1=get&data2=<?php echo $_GET['data2']; ?>"; 

      function refreshdiv(){ 

      // The XMLHttpRequest object 

      var xmlHttp; 
      try{ 
      xmlHttp=new XMLHttpRequest(); // Firefox, Opera 8.0+, Safari 
      } 
      catch (e){ 
      try{ 
      xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); // Internet Explorer 
      } 
      catch (e){ 
      try{ 
      xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); 
      } 
      catch (e){ 
      alert("Your browser does not support AJAX."); 
      return false; 
      } 
      } 
      } 

      // Timestamp for preventing IE caching the GET request 

      fetch_unix_timestamp = function() 
      { 
      return parseInt(new Date().getTime().toString().substring(0, 10)) 
      } 

      var timestamp = fetch_unix_timestamp(); 
      var nocacheurl = url+"?t="+timestamp; 

      // The code... 

      xmlHttp.onreadystatechange=function(){ 
      if(xmlHttp.readyState==4){ 
      document.getElementById(divid).innerHTML=xmlHttp.responseText; 
      setTimeout('refreshdiv()',seconds*1000); 
      } 
      } 
      xmlHttp.open("GET",nocacheurl,true); 
      xmlHttp.send(null); 
      } 

      // Start the refreshing process 

      var seconds; 
      window.onload = function startrefresh(){ 
      setTimeout('refreshdiv()',seconds*1000); 
      } 
      <!-- 
    refreshdiv(); 
    // --> 
    </script> 
    <script type="text/javascript"> 
    function initialize() { 
    var myLatLng = new google.maps.LatLng(<?php echo $cord_pos_start[0]; ?>); 
    var mapOptions = { 
     zoom: 15, 
     center: myLatLng, 
     mapTypeId: google.maps.MapTypeId.ROADMAP 
    }; 

    var map = new google.maps.Map(document.getElementById("map-canvas"), mapOptions); 

    var flightPlanCoordinates = [ 
     <?php echo implode('', $cords); ?> 
    ]; 

    var image_start = 'http://labs.google.com/ridefinder/images/mm_20_green.png'; 

    var marker = new google.maps.Marker({ 
     position: myLatLng, 
     map: map, 
     title:"Start", 
     icon: image_start  
    }); 

    var image_end = 'http://labs.google.com/ridefinder/images/mm_20_red.png'; 

    var endMarker = new google.maps.Marker({ 
     position: new google.maps.LatLng(<?php echo $cord_pos_end[0]; ?>), 
     map: map, 
     title:"Ende", 
     icon: image_end  
    }); 

    var flightPath = new google.maps.Polyline({ 
     path: flightPlanCoordinates, 
     strokeColor: "#FF0000", 
     strokeOpacity: 0.5, 
     strokeWeight: 5 
    }); 

    flightPath.setMap(map); 
    } 
    initialize(); 
    </script> 
    .... 

第一刷新我看看地图就可以了后一个刷新我的路径前得到这个错误：

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Alt\FIFA-Europaliga\content\show_runs.php on line 4

Notice: Undefined variable: cords in C:\xampp\htdocs\Alt\FIFA-Europaliga\content\show_runs.php on line 8

Notice: Undefined offset: 1 in C:\xampp\htdocs\Alt\FIFA-Europaliga\content\show_runs.php on line 9

Notice: Undefined variable: cords in C:\xampp\htdocs\Alt\FIFA-Europaliga\content\show_runs.php on line 11

有人能帮助我吗？

Answer 1

在运行查询之前，您必须连接到服务器并选择一个数据库。

$db = mysql_connect('hostname|ip', 'user', 'password'); 
mysql_select_db('database_name', $db); 
$result = mysql_query('SELECT ...', $db); 

如果查询失败，mysql_query将返回false，这将解释有关期待资源和获取布尔值的错误。

您得到的关于mysql_fetch_array需要资源的警告意味着查询失败。原因则可能无法包括：

1. 没有连接到正确的数据库
2. 数据库
3. 表上不存在缺少权限，被输错，等
4. 输入错误的字段，查询等。（错误的查询）

在所有情况下，都可以通过打印出mysql_error()来检查错误。