[RESOLVED] displaying images

mrbill501

Hi
I'm trying to make a small image gallery for a uni project. At the moment users can log in upload files etc but I cant get any of the files to display. Im storing the images in a blob in an sql table. I know this is frowned upon but Im a bit short of time now to go back and start again. My code for getting the images is:
<?php
set_magic_quotes_runtime(0);
ini_set("include_path","/home/fhlinux192/s/studentart.streamlinenettrial.co.uk/user/htdocs/includes/");
include ('pics.inc');
if($id) {
$connection = mysql_connect($host,$user,$password)
$db = mysql_select_db($database, $connection)
$query = "SELECT image, image_type FROM tbl_images WHERE image_id = $id";

$result = mysql_query($query);

$data = mysql_result($result,"image");
$type = mysql_result($result,"image_type");

Header( "Content-type: $type");
$data = stripslashes($data);
echo $data;

};
?>[/COLOR]

And the syntax for the images is: <img src="getdata.php3?id=334">

At the moment all I get is the white box with the cross in when I use these links.
Any help would be awesome.
Thanks
Andy

Jasp182

check out the [MAN]gd[/MAN] library.

If you have the image data in a string, you can use [MAN]imagecreatefromstring[/MAN] and then output it with one of the ImageXXX() functions (ex: [MAN]ImageJPEG[/MAN]). Your code suggests you have the mime-type stored in the db as well, so just select the corresponding functions. Currently, GD only supports JPEG, PNG, and BMP (at least the last time I checked that was true).

I've never used it before, but [MAN]ImageGD[/MAN] might work as well (and eliminate the need to switch/case the mime-type to decide what function to use). I haven't tested this theory yet though. =]

Oh, and you have to have GD compiled into your version of PHP

ClarkF1

Instead of if($id) try if(isset($_GET['id])) seeing as your parsing id through the URL

$id hasn't been declared anywhere. If you were to echo $id you'd probably get nothing.

You could use the following if you didn't want to type $_GET['id'] all the way through

if (isset($_GET['id'])) {
$id = $_GET['id'];
// rest of your code........

mrbill501

Ok, I've changed the code round so now it is:
if (isset($GET['$image_id'])) {
$id = $GET['$image_id'];
$connection = mysql_connect($host,$user,$password)
$db = mysql_select_db($database, $connection)
$query = "SELECT image, image_type FROM tbl_images WHERE image_id = $id";

$result = mysql_query($query);

$data = mysql_result($result,"image");
$type = mysql_result($result,"image_type");

Header( "Content-type: $type");
$data = stripslashes($data);
echo $data;

};
?>[/COLOR]

The image syntax is: ><img src="getdata.php3?image_id=334">

I'm still pretty new to this so I may come off as a little simple here but the images syntax looks at the getdata.php file which runs the select statement and looks for the image_id specified in the image code. Is that right? It does this through the if COLOR=Blue {
$id = $_GET['$image_id']; [/COLOR] code?
Does this make any sense??? I'd really like to be able to make this work and learn how and why it works rather than just doing it.
Thanks again.