Re: IF command
| From: | Casey | Date: | Fri, 19 Oct 2007 02:21:40 +0000 |
| Subject: | Re: IF command | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-263348@lists.php.net to get a copy of this message | ||
if (!in_array($component_reference, array(5,17)))
// ...
More compact.
On Oct 18, 2007, at 5:49 PM, "ron.php" <ron.php@actsministries.org> wrote:
When $component_reference is 5 or 19 I don't want the echo statement to output to the screen. What is the correct syntax? What I have below isn't working. Thanks for your help. Ron if ( ($component_reference != "5") OR ($component_reference != "19") ) { echo "<li><a href=\"index.php?request=" . $request . "\">" . $component_name . "</a>"; } --PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php