How to display code if it is not certain pages?

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Im trying to set up an if statement to display certain code, if the user is on a certain page. I’ve got the links and student work pages to work, but when I add the if not link and not student work I get an unexpected T_BOOLEAN_AND error. How can I fix this so I can list the pages I don’t want custom code for and use just page.php 

<?php 
            if (!is_page('links')) && (!is_page('student-work'))  {
                include(TEMPLATEPATH . '/pages/page.php');
            } elseif (is_page('links')) {
                query_posts('cat=2');
                include(TEMPLATEPATH . '/pages/links.php');
            } elseif (is_page('student-work')) {
                query_posts('cat=4');
                include(TEMPLATEPATH . '/pages/student-work.php');
            }
?>
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2 comments

  1. I just used this code and it works for me. Hopefully it solves your problem.

    if( ! is_page( array( 'links', 'student-work' ) ) ) { /* ... */ }

  2. This is really just PHP, not WP, but your initial if statement has an extra parenthesis after !is_page('links'), so the && is outside and PHP doesn’t know what to do with it. It’d be better to reorganize the block anyway, e.g.

    <?php 
        if (is_page('links')) {
            query_posts('cat=2');
            include(TEMPLATEPATH . '/pages/links.php');
        } elseif (is_page('student-work')) {
            query_posts('cat=4');
            include(TEMPLATEPATH . '/pages/student-work.php');
        }
        else  {
            include(TEMPLATEPATH . '/pages/page.php');
        }
?>

    In terms of WordPress, you’d probably be better off using page-specific template files, such as page-links.php or page-student-work.php. You could also use get_template_part() instead of includes to accomplish what that if/else block does in just one line if you need to keep this all in one file for some reason.