自動查詢MySQL並使用AJAX輸出結果

Question

我在Google地圖上自動顯示我的gps座標時遇到問題。當我實現地圖是靜態的時候一切正常，但現在我動態地嘗試它。這意味着當新的GPS座標將被保存到數據庫時，地圖應該自動刷新。

我在一個.php文件中實現了一切。爲了顯示與COORDS的地圖，用戶必須選擇一個條目，鏈接到一個條目是這樣的 - >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> 
    .... 

第一刷新我看看地圖就可以了後一個刷新我的路徑前得到這個錯誤：

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

有人能幫助我嗎？

Answer 1

在運行查詢之前，您必須連接到服務器並選擇一個數據庫。

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

如果查詢失敗，mysql_query將返回false，這將解釋有關期待資源和獲取布爾值的錯誤。

您得到的關於mysql_fetch_array需要資源的警告意味着查詢失敗。原因則可能無法包括：

1. 沒有連接到正確的數據庫
2. 數據庫
3. 表上不存在缺少權限，被輸錯，等
4. 輸入錯誤的字段，查詢等。（錯誤的查詢）

在所有情況下，都可以通過打印出mysql_error()來檢查錯誤。