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 (!$resultado['Imagem_data']) { ... } Sometimes testing a value against "" can cause problems because of non-appearing whitespace that may exist in the variable. SOMETIMES, not all the time, and I really couldn't tell you EXACTLY how PHP reacts to some of these conditions, because I believe it to be system-dependent if memory serves me (kind of like the differences between "\n" on Unix/Linux and Windows). Just play around with things, but I'd like to see the code you are using for this condition. Mike Frazer "Rodrigo Peres" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > List, > > I'm using the following code to retrieve image from Mysql. My problem is how > can I output another image if the given ID doesn't have an image on it??? I > had already inserted a blank gif in the database in order to use it, but > I've tried to check if ($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 > -- > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]