I comment it out but I got this error: Warning: Undefined variable: fileId in /home/narvaez/public_html/ddownloadfile.php on line 22 Could not get file list: You have an error in your SQL syntax near '' at line 1
Thanks for any help!. -----Original Message----- From: Sanduhr [mailto:[EMAIL PROTECTED]] Sent: Wednesday, February 20, 2002 6:58 PM To: [EMAIL PROTECTED] Subject: [PHP] Re: NULL value for variable that "global $fileId;" shouldn't be there in downloadfile.php "Teresa Narvaez" <[EMAIL PROTECTED]> schreef in bericht [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... : Hello, : I am runnig php 4.1.1. In the configuration, register_globals is : ON. : I have two programs and I want to pass the value from fileId from one.php to : downloadfile.php. However, in downloadfile.php $fileId is NULL. What am I : missing? Thanks, -Teresa : : : one.php : <td width="33%" bgcolor="#FFDCA8" height="21"> : <p style="margin-left: 10"> : <font face="Verdana" size="1"> : <a href="downloadfile.php?fileId=<?php echo $row["PicNum"]; ?>" > : Download Now : </a></font> : </td> : </tr> : : downloadfile.php : <? : global $fileId; : : if ( ! is_numeric($fileId) ) : die ("Invalid PictureNumber specified: ($fileId)"); : : ?> : : -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php