Re: get image from blob in mysql

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 (!$resultado['Imagem_data']) { ... } Sometimes testing a value against "" can cause problems because of non-appearing whitespace that may exist in the variable. SOMETIMES, not all the time, and I really couldn't tell you EXACTLY how PHP reacts to some of these conditions, because I believe it to be system-dependent if memory serves me (kind of like the differences between "\n" on Unix/Linux and Windows). Just play around with things, but I'd like to see the code you are using for this condition. Mike Frazer "Rodrigo Peres" <rperes@mac.com> wrote in message news:B8783A47.86EB%rperes@mac.com... > List, > > I'm using the following code to retrieve image from Mysql. My problem is how > can I output another image if the given ID doesn't have an image on it??? I > had already inserted a blank gif in the database in order to use it, but > I've tried to check if ($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 > -- > >