здравейте
Та въпроса ми е как да взема id-то на снимката от gallery.php във photos.php. Пробвах с $_GET[id] ама нещо не се получи :/
Код:
gallery.php
.........
$ImageDir ="images/";
$ImageThumb = $ImageDir . "thumbs/";
switch($_GET['get'])
{
default:
$limit = 12;// Set the limit of results per page
$page = $_GET['page'];// Sets the page, the default is ?page=
$totalrows = mysql_num_rows(mysql_query("SELECT `image_id` FROM `images`"));// Counts the total rows
if(empty($page))// If the page is empty
{
$page = '1';// Page is 1
};
$start = ($page-1)*$limit;
$start = round($start,0);// Sets the start
echo "<div id=\"imag\">";
$result = mysql_query("SELECT * FROM images ORDER BY image_id DESC LIMIT $start, $limit ") or die(mysql_error());
while ($r = mysql_fetch_array($result)) {
extract($r);
echo "<div id=\"imag2\">";
echo "<font color=\"#000\"> $image_caption </font>";
echo "<br>";
echo "<a href=\"photos.php?image=$r[image_id]\">";
echo "<img src=\"" . $ImageThumb . $image_id . ".jpg\" border=\"0\" id=\"imag3\" >";
echo "</a><br>";
echo "<font color=\"#000\">";
echo date('d.m.Y, H:i:s',$r[image_date]);
echo "</font>";
echo "</div>";
}
echo "</div>";
.......
Код:
photos.php
...
$getpic=mysql_query("SELECT * FROM `images` WHERE image_id='$id'") or die (mysql_error());
$rows=mysql_fetch_array($getpic) ;
extract($rows);
$image_filename="images/".$image_id.".jpg";
list ($width, $height, $type, $attr)=getimagesize($image_filename);
?>
<div id="body">
<h1> So how does it feel to be famous?</h1><br><br>
<p>Here is the picture you just uploaded to our servers:</p>
<img src="<?php echo $image_filename; ?>" align="left"<?php echo $attr; ?>>
<strong> <?php echo $image_caption; ?> </strong><br>
It is <?php echo $width; ?> pixels wide
and <?php echo $height; ?> pixels high.<br>
It was uploaded on <?php echo date('d.m.Y, H:i:s',$rows[image_date]); ?>
by <?php echo $image_username; ?>.
</div>
...Та въпроса ми е как да взема id-то на снимката от gallery.php във photos.php. Пробвах с $_GET[id] ама нещо не се получи :/
))