if($id) {
$id = $pet;

you first check for $id and then set it? 

> while(list($pet)=
> mysql_fetch_row($ret))
> print("<BR>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 = $pet (id also has a value of 3 in image_data)  why doesn't it
> show the image?
>
> if($id) {

if you want it for every row, there are missing { } with the while statement.

Op vrijdag 01 maart 2002 01:40, schreef Jennifer Downey:
> Hi all,
>
> I'm really stuck and I'm not asking anyone to re-write this just show me
> what is wrong, or explain why it wont work. It just seems logical that this
> should work.
>
> The first query will print the pet id in the browser.
>
> $query="SELECT  id FROM wt_users WHERE uid={$session["uid"]}";
> $ret = mysql_query($query);
> while(list($pet)=
> mysql_fetch_row($ret))
> print("<BR>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 = $pet (id also has a value of 3 in image_data)  why doesn't it
> show the image?
>
> if($id) {
>
>
> $id = $pet;
>     $query = "select bin_data,filetype from image_data where id=$id";
>
>   $result = mysql_query($query);
>
>     $data = mysql_query($result,0,"bin_data");
>     $type = mysql_query($result,0,"filetype");
>
>     Header( "Content-type: $type");
>     echo $data;
>
> };
> echo "<img src=\"petdata.php?id=$id\">";
>
> Thanks
> Jennifer Downey

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