Ok this is my code... It will found 7 data in my database.. But now my question is how could I show only the fift ones one a page and then put a link to the 3 other on a second page... This is the code.. I hope someone could take a few min to help me..
<?php
mysql_connect(localhost,twobig,841819519) or die("Impossible de se connecter a la database.<br>" . mysql_error());
mysql_select_db(twobig) or die("ON ne peut pas sélectionner de database.<br>" . mysql_error());
?>
<table width="75%" border="0" cellpadding="0">
<tr>
<td width="20%"></td>
<td width="40%">
<?php
$con = mysql_connect(localhost,twobig,841819519) or die("Impossible de se connecter a la database.<br>" . mysql_error());
$sql = "select from requete";
$result = mysql_db_query("twobig",$sql,$con);
echo $rows = mysql_num_rows($result);
?> Villas Trouvées</td>
<td width="40%">
<div align="right">Page suivante</div>
</td>
</tr>
</table><br>
<?php
/ Début de l'affichage de la requete /
$sql = "SELECT FROM requete";
$query = mysql_query($sql) or die("Erreur Database<br>" .mysql_error());
?> <table width="75%" border="0" cellpadding="0"> <?php
while($result = mysql_fetch_array($query)) {
$ID = stripslashes($result["ID"]);
$Nom = stripslashes($result["Nom"]);
$Personne = stripslashes($result["Personne"]);
$Description = stripslashes($result["Description"]);
$Images = stripslashes($result["Images"]);
?>
<tr>
<td width="20%"><?php echo "<img src=\""; echo "$Images"; echo "\">"; ?></td>
<td width="80%" valign=top>
<table width="100%" border="0" cellpadding="0">
<tr>
<td><font size="2" color="#330099"><?php print "<b>Reference : $ID </b>"; ?></font></td>
</tr>
<tr>
<td><font size="2" color="#003300"><?php print "<u>$Nom</u>, $Personne Personne(s)"; ?></font></td>
</tr>
<tr>
<td><font size="2" color="#330099"><?php print "$Description" ?></font></td>
</tr>
</table>
</td>
</tr>
<?php
} ?>
</table>
