Try doing echo stripslashes($data);
Or try setting the content type to text/plain - This should echo back
gobblygook (Just so you know that there IS image data there.
Also.. you have in another file right? If
you call petdata.php by itself, you should see just the image.
JD
-Original Mes
$query="SELECT id FROM wt_users WHERE uid=($session['uid'])";
Always print your queries if they do not work. I don't know if { } will
mess anything up since I've never used them. Your doublequotes in your
array will end your statement at [.
if( isset( $id ) ) { <-- use isset (http://www.ph
Hi Jennifer. Change {$session["uid"]} to {$session['uid']} --> double
quotes to single quotes. It's inside a set of single quotes already, so
you have to use double quotes.
/bsh/
Jennifer Downey wrote:
>Hi all,
>
>I'm really stuck and I'm not asking anyone to re-write this just show me what
if($id) {
$id = $pet;
you first check for $id and then set it?
> while(list($pet)=
> mysql_fetch_row($ret))
> print("your pet id is $pet");
>
> So if $pet will print the id from the wt_users (has a value of 3) and I
> assign $id = $pet (id also has a value of 3 in image_data) why doesn't it
>
