Hi list, I have some images stored in a blob in mysql. I've made this code to output them to html, but the problem is if there's no image at a given id the page tooks a long time to load and display the "broken image", there's a way to avoid this, I mean, there's a way to print a message or another image if there's none in mysql ???
<? $sql = "SELECT Imagem_data,Imagem_type FROM imagens WHERE CelebID='$celebID'"; $query = new Query($conexao); $query->executa($sql); $resultado = $query->dados(); mysql_close(); $imagem_banco = $resultado['Imagem_data']; $type = $resultado['Imagem_type']; if($imagem_banco != "") { HEADER("Content-type: $type"); echo($imagem_banco); } } ?> Thank's in advance 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]