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  • Can't get a display albums function to work... php [closed]

    - by Zhenia
    need your help with the code, please. I am trying to display an album from the database, but i just get some strange signs having no idea why... the signs are Albums Help me out if you know how to solve this problem. how can i maKe the code to display name of the album and amount of images in it? i am sure all is fine with the db. this is the function: function get_albums() { $albums = array(); // not always have got to put brackets $albums_query = "SELECT albums.album_id, albums.timestamp, albums.name, LEFT(albums.description, 50) as description, COUNT(images.image_id) as image_count FROM albums LEFT JOIN images ON albums.album_id = images.album_id WHERE albums.user_id = '{$_SESSION['user_id']}' GROUP BY albums.album_id"; $res = mysql_query($albums_query) or die(mysql_error().'<br>'.$albums_query); while($albums_row = mysql_fetch_assoc($res)){ $albums = array ( 'id' => $albums_row['album_id'], 'timestamp'=> $albums_row['timestamp'], 'name' => $albums_row['name'], 'description' => $albums_row['description'], 'count' =>$albums_row['image_count'] ); } return $albums; } and the other half of the code: <?php $albums = get_albums(); if(empty($albums)) { echo'<p>You don\'t have any albums</p>'; }else{ foreach($albums as $album){ echo'<p><a href="view_album.php?album_id=',$album['id'],'">',$album['name'],'</a>(',$album['count'],'images)<br /></p>'; } } ?>

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