Question

The code below will exclude the content type ‘name’ from being printed on a Page and Story type. Question, how can I do the opposite of this and print the content type name on only a Page and Story type?

    <?php
    if (!in_array($node->type, array('story', 'page'))) {
    print node_get_types('name', $node);
    }
    ?>

I removed the (!) and it works. How can I embed a css style into this scprit. I get error when I tried it this way. I just need the style to show on Page or Story types.

    <?php
    if (in_array($node->type, array('story', 'page'))) {
    <div class="mystyle">
    print node_get_types('name', $node);
    </div>
    }
    ?>

I got the css style in using ‘print’

    <?php
    if (in_array($node->type, array('story', 'page'))) {
    { print '<div class="mystyle"> ';
    print node_get_types('name', $node);
    }
    print '</div>';
    }
    ?>  

Answer 1

Assuming I understand the question correctly, just remove the not (!) operator, e.g:

<?php
if (in_array($node->type, array('story', 'page'))) {
     print node_get_types('name', $node);
}
?>