Am 2008-11-03 14:43:43, schrieb Anders Norrbring:
> I've been staring myself blind, so now I don't get anywhere, please do
> advice..
>
> I have a web page printed with PHP, in a table I need to display images that
> are stored in a SQL DB.
> Getting the images into variables isn't an issue at all, but how do I output
> it?
> This is what I want to accomplish:
>
> <tr><td>Current image:</td><td> - THE IMAGE HERE - </td></tr>
In the Website:
<tr><td>Current image:</td><td><img src="/?image=foo"></td></tr>
where I assume, you have a table called "imgstore" with three columgs of
"name", "type" and "image_data" and in the script:
if ($_GET['image'] != '') {
$connect="host=localhost dbname=imgstore user=foo password=bar";
$link=pg_connect($connect);
$query="SELECT image_data,type FROM images WHERE name=$_GET['image']";
list($image_data, $type=pg_query($query, $link);
header("Content-Type: image/$type; name=$_GET['image']");
header(2Content-Disposition: inline; filename=$_GET['image']");
echo $image_data;
pg_close($link);
exit();
}
where $type it for example "jpeg" or "png" and $image_data the blob you
fetched from your database.
ATTENTION: Please sanitize it sonce this is ONLY an EXAMPLE.
Thanks, Greetings and nice Day/Evening
Michelle Konzack
Systemadministrator
24V Electronic Engineer
Tamay Dogan Network
Debian GNU/Linux Consultant
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