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