. . "HERE img_id = .$id;$ret = mysql_query ($req) or die (mysql_error ());$col = mysql_fetch_row ($ret);if ( !$col[0] ){echo Id d'image inconnu;}else{header (Content-type: .$col[1]);echo $col[2];}}else{echo Mauvais id d'image;}?>if i use directly the show-image.php code there is another problem because of the header, i red on the web that there is a problem whith the header in a HTML/PHP page." . .