The code below doesn't show the test you mentioned.  By all means testing if
a variable = "" should work, but you may want to try something like

if (!$resultado['Imagem_data']) {

Sometimes testing a value against "" can cause problems because of
non-appearing whitespace that may exist in the variable.  SOMETIMES, not all
the time, and I really couldn't tell you EXACTLY how PHP reacts to some of
these conditions, because I believe it to be system-dependent if memory
serves me (kind of like the differences between "\n" on Unix/Linux and
Windows).  Just play around with things, but I'd like to see the code you
are using for this condition.

Mike Frazer

"Rodrigo Peres" <[EMAIL PROTECTED]> wrote in message
> List,
> I'm using the following code to retrieve image from Mysql. My problem is
> can I output another image if the given ID doesn't have an image on it???
> had already inserted a blank gif in the database in order to use it, but
> I've tried to check if ($resultado['Imagem_data'] == "") or  null and
> outputs the blank gif, but didn't work.
>     $conexao = new conexao();
>     $query = new Query($conexao);
>     $sql = "SELECT Imagem_data,Imagem_type FROM imagens WHERE
> CelebID='$celebID'";
>     $query->executa($sql);
>     $resultado = $query->dados();
>     $imagem_banco = $resultado['Imagem_data'];
>     $type = $resultado['Imagem_type'];
>     if($imagem_banco != "") {
>         HEADER("Content-type: $type");
>         echo($imagem_banco);
>     }
> Thank's
> Rodrigo
> --

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