[MySQL] formulaire combobox

bonjour j'ai un formulaire en PHP qui permet d'afficher un enregistrement d'une table mysql a partir d'un combobox
voila le formulaire combobox

Code:

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<?php 
$sql= "select device_id from assembly ";
$query = mysql_query($sql);
$numrow = mysql_num_rows($query);
echo '<form   action="updateform.php"  method="POST">';
echo "<select name=select1 action=updateform>";
 
for($i=0;$i<$numrow;$i++)
{
$champ1 = mysql_result($query,$i,'device_id');
echo "<option value=\"".$champ1."\">".$champ1."</option>";
}
echo "</select>";
echo'<INPUT type=submit value=valider name=ok><INPUT type=reset value=Annuler>';
echo'</form>';

et voila le formulaire PHP qui doit m'affiché l'enregistrement de la table selon la valeur choisie dans le combobox

Code:

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 if(isset($_POST['N']))      $N=$_POST['N'];
else      $N="";
 
if(isset($_POST['device_id']))      $device_id=$_POST['device_id'];
else      $device_id="";
 
if(isset($_POST['Tool_description']))      $Tool_description=$_POST['Tool_description'];
else      $Tool_description="";
 
if(isset($_POST['Tool_Area']))      $Tool_Area=$_POST['Tool_Area'];
else      $Tool_Area="";
 
if(isset($_POST['Tool_Location']))      $Tool_Location=$_POST['Tool_Location'];
else      $Tool_Location="";
 
if(isset($_POST['PDF']))      $PDF=$_POST['PDF'];
else      $PDF="";
 
 $query = "SELECT device_id,Tool_description,Tool_Area,Tool_Location,PDF  from assembly  WHERE  device_id='".$_POST["device_id"]."'";  
		$result = mysql_query($query) or die(mysql_error());
echo '<form method="POST" action="update.php">';
 
 
while ($row = mysql_fetch_row($result)) {  
 
 $device_id = $row[0];
 $Tool_description=$row[1];
 $Tool_Area=$row[2];
 $Tool_Location=$row[3];
 $PDF=$row[4];
 
 
 
echo "<br>device_id<input  name='device_id' size='50' value='$device_id'></input></br>";
echo "<br>Tool_description<input name='Tool_description' size='50' value=$Tool_description> </input></br>";
echo "<br>Tool_Area<input  name='Tool_Area' size='50' value='$Tool_Area'></input></br>";
echo "<br>Tool_Location<input name='Tool_Location'  size='50' value=$Tool_Location> </input></br>";
echo "<br>Fiche technique<input  name='PDF' size='50' value='$PDF'></input></br>";
 
echo'</form>';
 }
mysql_close();

mon probleme c'est qu'il m'affiche un formulaire vide avec l'erreur suivante
Notice: Undefined index: device_id in C:\Program Files\EasyPHP-5.3.8.0\www\Nouveau dossier\updateform.php on line 37