[ryanfern86goa]

hi there
I have created a table $query2="CREATE TABLE '$user1' ( id tinyint(3) unsigned not null auto_increment,
image blob not null, primary key(id))";
I have inserted quite a few pictures in it... when i try to display or fetch them i only get one picture dispalyed, thats the pic with id 1.heres my code

PHP Code:

<? session_start(); ?>
<?php
$user1=$_SESSION["user1"];
$pass1=$_SESSION["pass1"];
@mysql_connect("localhost", "abc", "abcd") or die("Can not connect to database: ".mysql_error());
@mysql_select_db('test') or die("Can not select the database: ".mysql_error());
$query = mysql_query('select image from $user1');
$row = mysql_fetch_array($query);
$content = $row['image'];
header('Content-type:image/jpg');
echo$content;
?>

i want to display all the pictures in that database..
Any help would be appreciated.. thanks

[wirehopper]

You need to add some code to indicate which image, by id, to display, then you need a second PHP file to loop through the ids and display the images.

You might want to leave the images in the file system and store an id, name, and caption in the database. Databases can't search images for content, so they don't add much value as a storage mechanism.

[ryanfern86goa]

hi hopper
thanks for ur reply.. was thinking of doing it by using while loop but really dont know how to go around it as i am new to php n mysql.. can you plz let me know about this
thanks

[wirehopper]

There are excellent examples here.

PHP: mysql_query - Manual