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: R> 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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
