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 . Ron

if ( ($component_reference != "5") OR ($component_reference != "19") ) {
echo "<li><a href=\"index.php?request=" . $request . "\">" .
$component_name . "</a>";
}

Hi, here is the code:

if ( ($component_reference != "5") AND ($component_reference != "19") ) {
echo "<li><a href=\"index.php?request=" . $request . "\">" .
$component_name . "</a>";
}


2007/10/19, ron.php <ron.php@actsministries.org>:
> When $component_reference is 5 or 19 I don't want the echo statement to
> output to the screen. What is the correct syntax? What I have below isn't
> working. Thanks for your . Ron
>
> if ( ($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
>
>

Are you sure that the value is a string? If its numeric, then try the checkwithout the quotes around the value


bastien> From: ron.php@actsministries.org> To: php-general@lists.php.net> Date: Thu, 18 Oct 2007 19:49:55 -0500> Subject: [php] IF command> > When $component_reference is 5 or 19 I don't want the echo statement to > output tothe screen. What is the correct syntax? What I have below isn't > working.Thanks for your . Ron> > if ( ($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>
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