Image Help!Urgent

phpmaverick

Working on a project as part of my final year, i need to get this to work quickly or i will flunk out of university!

What i need to do is get an image from a database and it to be placed on a specific location on the page. The variables used to call the image are x and y co-ordinates that are 'stored' in the url, they change with each click of the mouse and i need the image to change with them. i have attempted to do this myself using the below code:
<?
$x = $REQUEST['x'];
$y = $REQUEST['y'];

$db_name = 'localhost';
$table_name = 'Images';

$connection = @mysql_connect("localhost", "root", "*********") or die ("Couldn't connect.");

$db = @mysql_select_db($db_name, $connection) or die ("Couldn't select Database.");

$query = "SELECT * FROM '$table_name' WHERE x = '$xid' AND y = '$yid'";

$result = mysql_query ($query);

while ($row = mysql_fetch_array($result)) {
$image = $row['image'];

}

And then where the image should be i have: <IMG SRC="<? echo $image; ?>">

I'm sure this is me being an idiot but what can i say...i'm very new to php programing. Any help would be much appreciated!

RossC0

Hi,

Don't panic! you will need to use css to position your image.

So you may have to do some math but you need to output a style definition for the image - using your php variables.

Check out CSS and image

Check that out and you should be away!

phpmaverick

Thanks, that's helped with the positioning...i'm having trouble actually getting the image to appear at all though as well..any help with that? 🙂
Cheers
Maverick

RossC0

Whats it outputting? Need to check that $image is a valid image from the database?

So if the view source of the html page is:
<IMG SRC="">
Then its not setting $image properly - check there.

if its something like:

and its not displaying the image - then you need to ensure your pointing directly to the image from the directory the code is in.

So you might need to put a few: "../" notations so that it points to where the image is stored in relation to the php file.

Also I wouldn't use SELECT * notation if all you need is the image field - as its slightly wasteful (as you're not using everything else) - use SELECT image

Ross