Hi guys,
I need abit of your help. I have stored the images path in mysql database, so when I am using to get access to my site, something is like: www.mysite.com/link.php?user=test
It should look for the username row in the mysql database and prints for every images. But this time it can only print one image.
here's the code:
link.php
<?php
session_start();
define('DB_HOST', 'localhost');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_DATABASE', 'hf');
$errmsg_arr = array();
$errflag = false;
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
die("Unable to select database");
}
function clean($var)
{
return mysql_real_escape_string(strip_tags($var));
}
$username = clean($_GET['user']);
$pass = clean($_GET['pass']);
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
echo implode('<br />', $errmsg_arr);
} else {
$insert = array();
if(isset($_GET['user'])) {
$insert[] = 'username = \'' . clean($_GET['user']) .'\'';
}
if(isset($_GET['pass'])) {
$insert[] = 'pass = \'' . clean($_GET['pass']) . '\'';
}
if (count($insert)>0) {
$names = implode(',',$insert);
if($username) {
$sql = 'SELECT `id` FROM `images_list` WHERE `username` = "'.$username.'"';
$result1 = mysql_query($sql) or die('Error:<br />'. $sql .'<br />' . mysql_error());
$res = mysql_fetch_assoc($result1);
foreach($res as $row) {
echo '<p id="images"><img src="images.php?id='. $row['id'] .'" /></p>';
}
}
}
}
images.php
<?php
session_start();
define('DB_HOST', 'localhost');
define('DB_USER', 'mydbusername');
define('DB_PASSWORD', 'mydbpassword');
define('DB_DATABASE', 'mydbname');
$id = (int)$_GET['id'];
$errmsg_arr = array();
$errflag = false;
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
die("Unable to select database");
}
function clean($var)
{
return mysql_real_escape_string(strip_tags($var));
}
$qrytable1="SELECT images FROM images_list WHERE id=$id";
$result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
while ($row = mysql_fetch_array($result1)) {
$details = getimagesize($row['images']);
header ('Content-Type: ' . image_type_to_mime_type($details[2]));
echo file_get_contents($row['images']);
}
?>
any help would be much appreicated.