image and databases problems

OK, sorry for those who answered my first post, but i have decided to store
images in databases instead of as files on the server.
The problem i have is when i do a <img src="HTML/image.php?id=<?php echo
$row['id']; ?>">
, it does not show an image, just a small empty area,
CODE>> adding to a database
$imgFull =
imagecopyresized($newImg,$img,0,0,0,0,$fullWidth,$ fullHeight,$currentWidth,$currentHeight)
;
$imgSQL = "INSERT INTO images (id,image,width,height,image_type) VALUES
('$id' ,'{$newImg}', '$fullWidth' , '$fullHeight'
,'".$_SESSION["IMAGE_TYPE"]."') ";


CODE>> retrieving from database
$sql = "Select * from images where (id = '$id')";
$results = mysql_query($sql, $conn);
$row = mysql_fetch_array($results);
$image_type = $row["image_type"];
$image = $row["image"];
Header ("Content-type: ".$image_type);
echo $image;
mysql_close($conn);


if i just request the url (not embedded in a table and other HTML,) and
comment out the header function i get "Resource id #4" instead of all the
binary data i expected. I read that this was because this was a reference,
not the image

Thanks in advance
if you need any extra info, just ask

Jon

On Tue, 22 Jul 2003 19:56:48 +0100, Jon <jonchalk@hotmail.spam> wrote:

>OK, sorry for those who answered my first post, but i have decided to store
>images in databases instead of as files on the server.
>The problem i have is when i do a <img src="HTML/image.php?id=<?php echo
>$row['id']; ?>">
>, it does not show an image, just a small empty area,
> CODE>> adding to a database
>$imgFull =
>imagecopyresized($newImg,$img,0,0,0,0,$fullWidth, $fullHeight,$currentWidth,$currentHeight);

I'm assuming that all your variables are correctly defined, since there's no
information on them in this post.

>$imgSQL = "INSERT INTO images (id,image,width,height,image_type) VALUES
>('$id' ,'{$newImg}', '$fullWidth' , '$fullHeight'
>,'".$_SESSION["IMAGE_TYPE"]."') ";

That's not going to work... you've stored the image resource handle, not the
data representing the image.

And even if you had, you _absolutely need_ to run it through an escaping
function, e.g. mysql_escape_string.

http://uk.php.net/mysql_escape_string

There's no information here as to how you execute the query (if at all), or
whether you check for errors.

>CODE>> retrieving from database
>$sql = "Select * from images where (id = '$id')";

The brackets are superfluous.

>$results = mysql_query($sql, $conn);

You must check for errors. Do not ignore whether the return value is FALSE.

>$row = mysql_fetch_array($results);
>$image_type = $row["image_type"];
>$image = $row["image"];
>Header ("Content-type: ".$image_type);
>echo $image;
>mysql_close($conn);
>
>if i just request the url (not embedded in a table and other HTML,) and
>comment out the header function i get "Resource id #4" instead of all the
>binary data i expected. I read that this was because this was a reference,
>not the image

Yes - this is because you originally stored the image resource handle, NOT the
image data that it refers to.

Use imagepng, imagejpeg or whatever, and either write to a temporary file and
read it back in again for the data, or capture the output with the output
control functions.

Then escape that data, and store that in the database.

--
Andy Hassall (andy@andyh.co.uk) icq(5747695) (http://www.andyh.co.uk)
Space: disk usage analysis tool (http://www.andyhsoftware.co.uk/space)