Re: "include" question

Variables don't get parsed in single quotes, use double quotes. RClark wrote:

Hello all, I am passing a variable like so: <a href="link.php?foo=bar.php"> On the "link.php" page, I have this simple code: <?php $job = $_GET['foo']; echo "$job"; // for error checking include 'path/to/$job'; ?> The 'echo "$job";' statement works just fine, but the outbout for the include statement looks like this: bar.php Warning: Failed opening 'scripts/$job' for inclusion (include_path='.:/usr/local/lib/php') in /usr/local/www/data-dist/link.php on line 142 Can I not use a $variable in an include 'something.php '; statement? Thanks in advance, Ron Clark

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