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help using 'include'
Posted: Thu Jun 30, 2005 2:52 pm
by MasterCephus
I am trying to use an include function in my code, but I keep getting an error.
The problem is that I need to also send a URL wrapper.
For instance, the file is 'code.php', but I need it to be invoked as 'code.php?arg=x'
Is there anyway to do this?
I tried to follow the example from
here, but it doesn't seem to work.
Posted: Thu Jun 30, 2005 2:55 pm
by Chris Corbyn
include() is a filesystem level function.
Those query strings are only parsed when read via a web server so include ('
http://localhost/~username/file.php?page=blah');
Posted: Thu Jun 30, 2005 2:56 pm
by Burrito
you can't do that, you'll either have to just use the url params on the parent page, or set some right before you include the page...
Posted: Thu Jun 30, 2005 2:57 pm
by djot
guess, browser called file is:
index.php?arg_x=whatever
index.php:
test.inc.php:
Code: Select all
<?php
if (isset($_GET['arg_x'])) { $arg_x=$_GET['arg_x']; } // You may put this and the next line ...
else { $arg_x=''; } // ... into file index.php also.
if ($arg_x=='whatever')
{
//do something
}
else
{
// do something else
}
?>
Posted: Thu Jun 30, 2005 9:10 pm
by MasterCephus
Well, if I can only use the 'include("code.php")'
Is there a way I can send a variable to that page? like send it a $variable?
and if the variable is "" do something and if it's "x" do something else?
Posted: Thu Jun 30, 2005 11:32 pm
by Burrito
do what I told you to do.
you can either use the $_GET[] array that is passed from the url or you can set your own above the include:
ex:
Code: Select all
$_GETї'whatever'] = "e;bob"e;;
include("e;yourpage.php"e;);