Invité
Invité n'ayant pas de compte PHPfrance
05 déc. 2006, 17:29
Voila les codes:
config.php
=======
$date = getdate();
$year = $date[year];
$mois = $date[mon];
$jour = $date[mday];
$FormatDate = "$jour/$mois/$year";
Affichage de l'image du jour:
==================
<?php
include ("config.php");
if(isset($_POST['img'])) $img = $_POST['img'];
else $img = '';
if(isset($_POST['url'])) $url = $_POST['url'];
else $url = '';
$date = date("d/m/Y");
$connection = mysql_connect("$dbhost", "$dblogin", "$dbpassword") or die ($ErrorConnection);
$db = mysql_select_db($dbname, $connection) or die ($ErrorDBase);
$select = "SELECT * FROM ".$gallery_table." WHERE date='".$FormatDate."'";
$result = mysql_query($select,$connection) or die ($ErrorSelect);
$total = mysql_num_rows($result);
if($total==0)
{
echo $ErrorMessage2;
}
else
{
while ($data = mysql_fetch_array($result)) {
$img = $data['img'];
?>
<center>
<div>
<IMG onmouseover=pick(this) style="FILTER: alpha(opacity=100) gray()" onmouseout=unpick(this) src="<?php echo "$img"; ?>" width="120" height="120" border="0" class="imglink">
</center>
</div>
<div>More >>></div>
<?php
}
}
mysql_free_result ($result);
mysql_close ();
?>
</center>
insertion de l'image avec sa date:
=====================
<select name="jour" size=1>
<option selected value="1">01</option>
<option value="2">02</option>
<option value="3">03</option>
<select name="mois" size=1>
<option selected value="1">01</option>
<option value="2">02</option>
<option value="3">03</option>
<select name="year" size=1>
<option selected value="2006">2006</option>
$sql = "INSERT INTO $gallery_table VALUES ('','$jour/$mois/$year','$img','$url')";
$result = mysql_query($sql,$connection) or die ($ErrorInsert);
et dans la base de donnee j'ai ce resultat:
INSERT INTO `gallery_table` VALUES (14, '4/12/2006', 'logo_phpfrance10.gif', 'images/logo_phpfrance10.gif');
J'ai mis juste les champs ou j'ai doute qu'il y ait une erreur.
merci