在删除PDO PHP后填充表

Question

当用户登录时，他将转到一个有表的页面，该表在每行上都填充了delete链接。如果我点击删除链接，它将删除记录，但当它刷新页面时，表将不会加载此消息。

Warning: mysql_query(): Access denied for user ''@'host' (using password: NO)
Warning: mysql_query(): A link to the server could not be established in 
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given

为什么它会让我第一次又一次地填充表呢?这是我的删除代码。

$id_actividades = $_GET['idactividades'];
$stmt = $dbh->prepare("DELETE FROM guaynabodb.actividades WHERE idactividades=:id_actividades");
$stmt -> bindParam(':id_actividades', $id_actividades);
     try{
      $stmt->execute();
     }
     catch (PDOException $ex) {
    echo $ex->getMessage(),'Cannot delete',"n";
      }
      header('Location: actividades.php');//To redirect
      exit;

这是我拥有表格的页面的代码。它加载后，你登录，但如果我删除一个记录，它不会加载。

  include('../includes/dbschema.php');
      $stmt = $dbh->prepare("SELECT * FROM actividades");
          $stmt->execute();
     print " <h1 id="h2title">Calendario de Actividades</h1><br/><br/>";//Print the title header
           echo "<table id="premiacionguaynabo"> <tr> <th> No. </th> <th> Fecha </th> <th> Torneo </th> <th> Lugar </th> <th> Organizador </th> <th> Opciones </th> </tr>"; //The table and the headers are created  
             $tno = 0;
              $result = $stmt->fetchall(PDO::FETCH_ASSOC);
              foreach($result as $line){
                $tno = $tno + 1;
     $id = $line["idactividades"];
    print "<tr class="alt2">"; 
    print "<td id="idtorneo">  $tno  </td>";
    print "<td class="fechatorneo"> " . $line['fecha_inicial'] . " al " . $line['fecha_final'] .  "</td>";
    print "<td> <a id="plinks"  href="$picture" rel="lightbox" target="_top" title="Flyer del Torneo"> " . $line['ntorneo'] . " </a></td>";
    print "<td>" . $line['ltorneo'] . "</td>"; 
    print "<td>" . $line['otorneo'] . "</td>"; 
    print "<td id="idtorneo"> <a id="udlinks"  href="uactividades.php?idactividades=$id">  Edit  </a>   <a id="udlinks" onclick="return confirmDelete()"  href="dactividades.php?idactividades=$id">  Delete  </a></td>";
    print "</tr>"; 
        }
    print "</table>"; 

Answer 1

错误信息提到mysql_query()函数，该函数在您发布的代码中不存在。
所以，要么这些函数被意外地留在代码中，要么你只是发布了错误的代码。
请阅读您的错误信息。它们内容丰富。