jofftan

Not exactly. Ok now it's calling the same page within the link so it work I don't know if it's secure but it work. [code] while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpo.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; }[/code] But now I got this error to fix [quote]Notice: Undefined variable: id_team in /var/www/vhosts/virochempharma.com/httpdocs/corpo.php on line 106 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/virochempharma.com/httpdocs/corpo.php on line 109[/quote]

No I'm sorry I wasn't clear enough. The links are in the dropdown of team members in the middle of the page.

Here's the link [url=http://www.virochempharma.com/corpo.php]http://www.virochempharma.com/corpo.php[/url] When you get to this page I'm echoing the first entry of the DB. If you select another entry from the dropdown it call a second page. I want all that operation on a single page. I don't know if it's possible ?

Hi, I'm looking for a way to choose data content from a dropdown menu, linked to a MySQL DB, and echo it on that same page. The problem is when you get to that page for the first time I need to display the first entry of the data by default. Now I have it but with two php files, I want this on the same page. The way it is right now is Page 1 ; [code] <?php $c = mysql_connect ($host,$usager,$Mpasse); mysql_select_db ('db', $c); if (!$c) {   die('Impossible de se connecter : ' .  error_reporting(E_ALL)); }   $req = "SELECT * FROM Team"; $resultat = mysql_query($req); echo '<ul id="subnavlist">';   while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpoTeam.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; }     echo "</li>"; ?> [/code] Page 2; [code] <?php $_POST['id_team'];     include 'connexion.php'; $c = mysql_connect ($host,$usager,$Mpasse); mysql_select_db ('db', $c); if (!$c) {   die('Impossible de se connecter : ' . mysql_error()); }   $req = "SELECT * FROM Team"; $resultat = mysql_query($req); echo '<ul id="subnavlist">';   while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpoTeam.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; }     echo "</li>"; ?> </li> </ul> </div> <div> <?php           ini_set('display_errors', 1); error_reporting(E_ALL);             $req = "SELECT * FROM Team WHERE id_team = ".$id_team; //                        $resultat = mysql_query ("SELECT Titre, description, name FROM Team WHERE id_team = '".urldecode($_GET['id_team'])."'");                     $resultat = mysql_query($req);                     $row=mysql_fetch_array($resultat);                     $name = $row['name'];                     $Titre = $row['Titre'];                     $description = nl2br($row['description']);                                         echo "<br>";                     echo "<br>";                                  echo '<h4>';                     echo $name;                     echo '<br>';                     echo $Titre;                     echo '</h4>';                     echo '<p>';                     echo $description;                     echo '</p>';                     ?> [/code] Thank you