Въпрос относно галерия и база данни

[Knoppix_]

Здравейте,

имам следният код :

<?php

$hostname = 'localhost';
$username = '***';
$password = '***';
$database = 'shikomed_pic';


mysql_connect($hostname, $username, $password) || die(mysql_error());
mysql_select_db($database);
mysql_query ("SET NAMES 'UTF-8'");

$sql = "SELECT * FROM pictures order by id";

if(!$result = mysql_query($sql)) {
die(mysql_error());
}


$img=$row['pic_name'];
$image = '<img src="projects/'.$img.'" width="90" height="70" />';



while($row = mysql_fetch_array($result)) {
echo '<div>';
echo $image;
echo '</div>';
}
?>

Идеята е да взима от базата данни снимките и да ги отпечатва само че не ми се визуализарат снимките т.е седи ми празно място.

Таблицата е това

CREATE TABLE IF NOT EXISTS `pictures` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`pic_name` varchar(255) CHARACTER SET utf8 NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

 

[nom3r1]

Предполагам е това;

<?php
			$host = "host";
			$user = "user";
			$pass = "pass";
			$name = "name";
			$connect = mysql_connect($host, $user, $pass)or die(mysql_error());
			mysql_select_db($name, $connect)or die(mysql_error());
			
			mysql_query ("SET NAMES 'UTF-8'"); 
			$query = mysql_query("SELECT * FROM `pictures` ORDER BY id DESC") or die(mysql_error());
            while($row = mysql_fetch_array($query)){
            $snimka = $row['pic_name'];
            echo "<img src=\"projects/$snimka\" width=\"90\" height=\"70\" />";
   }
?>

Или греша..

 

[vesku]

<?php
$hostname = 'localhost';
$username = '***';
$password = '***';
$database = 'shikomed_pic'
mysql_connect($hostname, $username, $password) || die(mysql_error());
mysql_select_db($database);
mysql_query ("SET NAMES 'UTF-8'");
$sql = "SELECT * FROM pictures order by id";
if(!$result = mysql_query($sql)) {
die(mysql_error());
}
while($row = mysql_fetch_array($result)) {
echo '<div>';
echo '<img src="projects/'.$row['pic_name'].'" width="90" height="70" />';
echo '</div>';
}
?>