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.


Jennifer Downey wrote:

>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);
>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\">";
>Jennifer Downey


/---------------------------------------------=[ BILLY S HALSEY ]=--\
| Member of Technical Staff - Sun Microsystems, Inc.                |
| ESP Solaris Software - Software Problem Resolution Team           |
| "No day but today." - J. Larson, RENT         x55403/858-526-9403 |
\--=[ [EMAIL PROTECTED] ]=------------------------------------------------/

PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Reply via email to