Re: Re: help with Header call

David, I hope that Robbie was able to solve your problem. But, if not, try isolating the variables from the text in the header. e.g. header("Location: ".$relative."/...?II=".$II.""); Hugh ----- Original Message ----- From: "David Robley" <robleyd@paradise.net.nz> To: <php-general@lists.php.net> Sent: Monday, February 18, 2002 6:18 PM Subject: [PHP] Re: help with Header call > In article <OE4944wo0T0jd3VBQQg000061bc@hotmail.com>, > phpsqllists@hotmail.com says... > > Hello everyone, > > > > This is my first post to this list, so I hope that someone is awake enough to help me out. > > > > I am new to PHP and I have a file that uses a command "header". It isn't working like it is supposed to but I can't seem to figure out why. Below is the last couple lines of the php file that is giving me trouble. > > > > while ($row=mysql_fetch_row($result)) { > > $II=$row[0]; > > } > > > > Header("Location: $Relative/admin/upload.php?II=$II"); > > > > Basically all it does when the whole file is called is return page does not exist. The file that it refers to is there and the Var "$Relative" is used in many other scripts with no probs so I don't think that is it. Is passing the Var "$II" as part of the location incorrect? > > > > Any help is welcomed. > > > > If you need to see more of the script to understand what I am talking about, I will post it. > > > > TIA > > -Robbie > > > > Try echoing the value of $II before the calling the header function; also > try echoing the header parameter to see that it is what you expect it > should be. Of course you might want to comment out the actual header line > as you do the echo tests, as the output from the echos will cause an > error from header() > > -- > David Robley > Temporary Kiwi! > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >