Thanks! Works perfect with double quotes! RC
-----Original Message----- From: Tom Rogers [mailto:[EMAIL PROTECTED]] Sent: Thursday, December 12, 2002 8:21 AM To: Ronald Clark Cc: [EMAIL PROTECTED] Subject: Re: [PHP] "include" question Hi, Friday, December 13, 2002, 12:07:05 AM, you wrote: R> Hello all, R> I am passing a variable like so: R> <a href="link.php?foo=bar.php"> R> On the "link.php" page, I have this simple code: R> <?php R> $job = $_GET['foo']; R> echo "$job"; // for error checking R> include 'path/to/$job'; ?>> R> The 'echo "$job";' statement works just fine, but the outbout for the R> include statement looks like this: bar.php R> Warning: Failed opening 'scripts/$job' for inclusion R> (include_path='.:/usr/local/lib/php') in /usr/local/www/data-dist/link.php R> on line 142 R> Can I not use a $variable in an include 'something.php '; statement? R> Thanks in advance, R> Ron Clark You have to use double quotes like: include "path/to/$job" or add like this: include 'path/to/'.$job -- regards, Tom CONFIDENTIALITY NOTICE: ************************************************************************ The information contained in this ELECTRONIC MAIL transmission is confidential. It may also be privileged work product or proprietary information. This information is intended for the exclusive use of the addressee(s). If you are not the intended recipient, you are hereby notified that any use, disclosure, dissemination, distribution [other than to the addressee(s)], copying or taking of any action because of this information is strictly prohibited. ************************************************************************
