[Content on the same page of the query]

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]

[Content on the same page of the query]

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

[Content on the same page of the query]

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 ?

[Content on the same page of the query]

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