First, could you be more specific? What isn't working? What error messages
are you getting? (I assume you are echoing them, because if not, you should
be... mysql_query($query) or die (mysql_error());

Second, this could be a problem. I think you need single quotes around the
variable here:

    $query = "select bin_data, filetype from image_data where
id=------->>>>>>>>'$id'"; <<<<-----

Beyond that, I can't really help much. Never stored image data in a db
before, so I can't tell you if anything else might be amiss. Good luck.

"Jennifer Downey" <[EMAIL PROTECTED]> wrote in message
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

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