What is wrong with my table? It will not display DB

Question

I am new when It comes to PHP. I've been having issues displaying my table from mysql in my site. I found this code that doesn't seem to work. Can someone please tell me what's wrong the code? Or suggest a better way to output mysql to an html table.

<?php $username="user";$password="password";$database="database";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM tablename";$result=mysql_query($query);
$num=mysql_numrows($result);mysql_close();?>

               <div class="panel-body">
                   <table class="table table-bordered table-striped mb-none" id="datatable-tabletools" data-swf-path="assets/vendor/jquery-datatables/extras/TableTools/swf/copy_csv_xls_pdf.swf">
                       <thead>
                       <tr>
                           <th>id</th>
                           <th>Value1</th>
                           <th>Value2</th>
                           <th>Value3</th>
                           <th>Value4</th>
                           <th>Value5</th>
                           <th>Value6</th>
                           <th>Value7</th>
                           <th>Value8</th>
                       </tr>
                       </thead>
                       <?php$i=0;while ($i < $num)
                       $f1=mysql_result($result,$i,"id");
                           $f2=mysql_result($result,$i,"v1");
                           $f3=mysql_result($result,$i,"v2");
                           $f4=mysql_result($result,$i,"v3");
                           $f5=mysql_result($result,$i,"v4");
                            $f6=mysql_result($result,$i,"v5");
                            $f7=mysql_result($result,$i,"v6");
                            $f8=mysql_result($result,$i,"v7");
                            $f9=mysql_result($result,$i,"v8"); ?>
                       <tbody>

                       <tr class="gradeX">
                           <td><?php echo $f1; ?></td>
                           <td><?php echo $f2; ?></td>
                           <td><?php echo $f3; ?></td>
                           <td><?php echo $f4; ?></td>
                           <td><?php echo $f5; ?></td>
                           <td><?php echo $f6; ?></td>
                           <td><?php echo $f7; ?></td>
                           <td><?php echo $f8; ?></td>
                           <td><?php echo $f9; ?></td>
                       </tr>
                       </tbody>
                   </table>
                   <?php$i++;}?>
               </div> 

Answer 1

You do this:

$num=mysql_numrows($result);mysql_close();?>
                            ^^^^^^^^^^^^^

Before you do this:

                   <?php$i=0;while ($i < $num)
                   $f1=mysql_result($result,$i,"id");
                       ^^^^^^^^^^^^^^^^^

Can't fetch anything from your query result, since you killed the connection that the results would be coming from.

And note that the mysql_*() functions are obsolete and deprecated. You should NOT be using them in any new code.

Answer 2

In case u're ok with a fresh start using, here's the code to connect to the DB :

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "";

try 
{
    $db = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    echo "Connected successfully" . "<br /> \n";

}
catch (PDOException $e) 
{
    echo "Can't connect to the server !: " . $e->getMessage() . "<br />";
    die();
} 

 try 
 {  
      $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

      $sql = "";
      $stmt = $db->query($sql);
      $array_res = $stmt->fetchAll();

} 
catch (Exception $e) 
{
   echo "Failed: " . $e->getMessage();
}
?>

U'll just have to edit ur personnal infos on top & $sql. Then ure free to display it, all explained in the link below

PHP.net pdo::query