Thanks!  Works perfect with double quotes!


-----Original Message-----
From: Tom Rogers [mailto:[EMAIL PROTECTED]] 
Sent: Thursday, December 12, 2002 8:21 AM
To: Ronald Clark
Subject: Re: [PHP] "include" question


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
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




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