I created my pict field with a VAR(32) Type in wich I put the name of the image that stant on the same folder of my php fileé
Now i'm trying to get that iamge to show but it doesn't.
Can somebody have a look at this code and tell me where did I go wrong.
Please !
...
mysql_select_db ( "fiche", $connexion );
$result = mysql_query ("SELECT * FROM FrCandidat", $connexion);
$row = mysql_fetch_array($result);
$prenom = $row['prenom'];
$nom = $row['nom'];
$email = $row['email'];
$genre = $row['genre'];
$phone = $row['phone'];
$pict = $row['pict'];
$acces = $row['acces'];
$poste = $row['poste'];
$fax = $row['fax'];
$pratique = $row['pratique'];
$admission = $row['admission'];
$formation = $row['formation'];
$association = nl2br($row['association']);
?>
<div id="hdr">
<img class="img" src="<?= $pict ?>" alt="" width="144" height="144" border="0"><br/>
<img src="imagesFiche/logoS.gif" alt="" width="144" height="34" border="0">
<p class="Prenom"><?php echo $prenom ?> <span class="Nom"><?php echo $nom ?></span></p>
<p><span class="Av"><?php echo $genre ?></span></p>
<div id="Tel">
Téléphone<br>
Accès direct<br>
Télécopieur<br>
Courriel</div>
<div id="Tel2">
<?php echo $phone ?> <br>
<?php echo $acces ?> poste <?php echo $poste ?><br>
<?php echo $fax ?><br>
<?php echo "<a href='mailto:$email'>$email</a>"; ?></div>
</div>
...