display all images from database

hwoarang699

trying to display all image from database. type of my image is long blob.

but on my gallery.php it keep printing the 1st image in database.

i have a image.php file

<?php
    session_start();
    include("connect.php");

$user_id = $_SESSION['user_id']; 

$image = mysql_query("SELECT * FROM image WHERE user_id = $user_id ");
$image = mysql_fetch_assoc($image); //get access to image table
$image = $image['image'];
echo base64_decode($image); 
?>

i also have a gallery.php where iam trying to display all different images of user

$id = $_SESSION['user_id'];
$sql = "SELECT * FROM image WHERE user_id = $id";
	$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result))
	{
	<img src="image.php?id='.$row['image_id'].'" />
}

Weedpacket

Could it be because you're asking image.php to just get an image from the gallery based on "user_id" and not using the image ID that gallery.php is passing in $_GET['id'] to look the image by by its "image_id"?

hwoarang699

Notice: Undefined index: id in C:\xampp\htdocs\login_test\image.php on line 6

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\login_test\image.php on line 9

i tryed changing

$user_id = $_GET['id'];

and it give me two errors.

Weedpacket

Some of that is because you're using the wrong quotes in gallery.php (look at the syntax highlighting in your post, and look at the value of the src attribute for the image in your page).

i tryed changing [code=php wrote:
$user_id = $_GET['id']; [/code]

Why would you do that? You want $user_id to be the user ID, not the image ID, don't you?

hwoarang699

sorry, i just started learn about php some days ago. i dont understant how can i fix this.

so the problem is in image.php file where it say?

$user_id = $_GET['id'];

i should change it to $_GET['user_id']?