Var not affected on mysql - php

Question

Good morning all !
I have a problem with a var in my code :
<?Php
error_reporting(E_ALL);
//Si IP correspond pas au PC autorisé, on retourne au debut.
include "./_protection.php";
$data='';
$rep='';
$requete='';
$val = '';
$Reference= '';
echo '<h2><u>Commander des pièces</u></h2><br>';
/*****************************************************************************************************
* AJOUT DE PIECE DANS LA BASE DE DONNEES
*****************************************************************************************************/
if(#$_POST['action'] == "ajouter")
{
if(!empty($_POST['choix'])){
foreach($_POST['choix'] as $val){
$Reference.="$val,";
}
//Injection dans la base SQL
include "./../connexion.inc";
mysql_query("INSERT INTO `tbl_commandes` ( `id`, `Eleve`, `Reference`, `Quantite`, `Date`, `Statut` )
VALUES ( '', '".$_POST['Eleve']."', '".$_POST['Reference']."', '".$_POST['Quantite']."', '".$_POST['Date']."', '".$_POST['Statut']."' )");
mysql_close();
}
}
/*****************************************************************************************************
* FORMULAIRE POUR L'AJOUT DE PIECE
*****************************************************************************************************/
echo '
<fieldset>
<legend>Commander des Pièces</legend>
<FORM action="./?page=commande_piece_full_php" method="POST">
<p align="left">
<H2>Pièces :</h2>
<br>';
echo $Reference;
include "./../connexion.inc";
$rep = mysql_query("SELECT id_categorie_piece, categorie_piece FROM tbl_categorie_piece ORDER BY categorie_piece");
mysql_close();
include "./../connexion.inc";
while($data = mysql_fetch_array($rep))
{
echo $data['categorie_piece'];
echo '<br>';
$requete = $data['id_categorie_piece'];
$reponse = mysql_query("SELECT id_piece, piece FROM tbl_piece WHERE idr_categorie_piece='$requete' ORDER BY piece");
while($donnees = mysql_fetch_array($reponse)){
?> <input type="checkbox" name="choix[]" value="<? echo $donnees['id_piece'] ?>"><? echo $donnees['piece'] ?><br> <?
}
echo '<br>';
}
mysql_close();
echo '
Elève :
<select name="Eleve">';
include "./../connexion.inc";
$reponse2 = mysql_query("SELECT id_client, client FROM tbl_client ORDER BY client");
while($donnees2 = mysql_fetch_array($reponse2))
{
?> <option value="<? echo $donnees2['client'] ?>"><? echo $donnees2['client'] ?></option> <?
}
mysql_close();
echo '
</select>
<br>
Date souhaitée : <input type="textbox" name="Date">
<br>
Quantité : <input type="textbox" name="Quantite">
<br>
<input type="submit" value="ajouter">
<input type="hidden" name="action" value="ajouter">
</form>
</fieldset>';
?>
my var $reference is Ok, i can echo $reference; and get the content rightly
BUT when i wanna send $reference to my mysql table .. that doesn't send it
When i submit, i get this notice from the parser : Notice:
Undefined index: Reference in /home/bddstock/www/admin/commande_piece_full_php.php on line 27
but when it is submited i get the right content in my echo line under PIECES
so i really don't understand ..
can someone help me ?

Answer 1

Change your insert statement to
mysql_query("INSERT INTO `tbl_commandes` ( `id`, `Eleve`, `Reference`,
`Quantite`, `Date`, `Statut` )
VALUES ( '', '".$_POST['Eleve']."', '".$Reference."', '".$_POST['Quantite']."', '".$_POST['Date']."', '".$_POST['Statut']."' )");
i.e change $_POST['Reference'] to $Reference as you have not posted this field and it seems you derive it from looping $_POST['choix'].
Note : use mysqli instead of mysql and refer How can I prevent SQL injection in PHP?