note 27953 added to function.header

<?php mysql_connect($hostName, $userName, $password) or die ("can't connect¡C"); mysql_select_db($databaseName) or die("can't select DB"); $result = mysql_query($SQL = "SELECT * FROM pronotedetphoto WHERE po='$po' and id = '$id'") or die("MYSQL detect error¡G " . mysql_error() . "<br><br>" . "SQL is¡G $SQL<br><br>"); if ($result_ar = mysql_fetch_array($result)) { header("Content-type: image/jpeg"); echo $result_ar['file_data']; } else { die("this id doesn't contain image"); } ?> i can show the jpeg file on this php file, but i can't show any text. i know that i use header("Content-type: image/jpeg"); so i can't show any text on it. Also this php file is called by another link, <?php //callphoto.php printf("<td width=29%% rowspan=19%%><A HREF=\"ch15ex06.php?id=%s&pnno=%s\">see photo</A></td>",$row->id,$key); ?> if i want to show the jpeg file on the callphoto.php directly, what can i do? -- http://www.php.net/manual/en/function.header.php http://master.php.net/manage/user-notes.php?action=edit+27953 http://master.php.net/manage/user-notes.php?action=delete+27953 http://master.php.net/manage/user-notes.php?action=reject+27953