On Mon, January 22, 2007 10:11 pm, Aaron Axelsen wrote:
> I'm trying to figure out what the desired behavior is of using the
> return function to bail out of an include page.
>
> I did some testing, and this is what I concluded.
>
> First, I created the following file:
>
> <?php
> if (defined('TEST_LOADED')) {
> return;
The behaviour of return in the main script scope changed from, errr,
PHP 3 to PHP 4, so I'm not sure it's a good idea to rely on it.
> }
> define('TEST_LOADED',true);
> echo "blah blah blah blah<br/>";
> ?>
>
> I then called it as follows:
> include('test.php');
> include('test.php');
> include('test.php');
>
> The output is:
> blah blah blah blah
>
> Second, I changed the test.php file to be the following:
>
> <?php
> if (defined('TEST_LOADED')) {
> return;
> }
> define('TEST_LOADED',true);
> echo "blah blah blah blah<br/>";
>
> function myFunc($test) {
>
> }
> ?>
>
> When I load the page now, it throws the following error: PHP Fatal
> error: Cannot redeclare myfunc()
>
> It appears that if there are functions in the include page that you
> can't use return to bail out. What is the desired functionality in
> this
> case? Is this a bug in how php handles it? or was return never
> designed
> to be used this way?
You *could* do more like this:
if (!defined('TEST_LOADED')){
define('TEST_LOADED', true);
function my_func() { return true; }
}
This should work, I think...
You also could consider just using http://php.net/include_once and let
PHP handle this.
Or, you could actually architect your application to not be including
things so willy-nilly that you don't even know what the [bleep] you've
included... :-) :-) :-)
To be blunt, though, code like this usually IS a sign that the
application is a bit dis-organized, pulling in include files with too
little planning.
--
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Know what I want?
I want you to buy a CD from some starving artist.
http://cdbaby.com/browse/from/lynch
Yeah, I get a buck. So?
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