Passing variables errors in a nasty way! Is it PHP settings?
Posted: Wed Jul 16, 2003 8:28 am
Hello all, this is my first post.
I've installed many times php, mysql and apache (1.3x and 2.x) on my win2k home machine so I can do my dev testing at home before screwin up my server
On my original installation I managed to get it all down well with a foxserv installation but something went wrong, I uninstalled, re-installed (many times) and it kept erroring..
Here is the problem:
I have been using PHP to create a tables based page that has a menu on the left (which is included into a layout) and the content on the right (which is included into the same layout).
The way I do it is to pass a variable from my menu to the layout page which then places the content on the right according to the variable passed.
I presume this is a common way of doing things, as I'm a newby I was greatly impressed with this simple feature alone to make managing large sites simpler...
OK, back to the problem...
Now I've re-installed the software (apache/mysql/php) it all works fine except the ability to pass the variable and accept it to re-generate the page with the appropriate content.
Here is the code I use in my menu:
Here is the code I use in my layout that accepts the $id variable. Note: it's placed in a table's cell for layout purposes...
I haven't included <? and ?> in any of these examples, please imagine they exist
So that always worked fine.. the menu code would send the $id to the layout area which would turn it into whatever menu item i've clicked...
NOW THOUGH.. this is the error I get:
Warning: Failed opening '' for inclusion (include_path='.;C:/phpdev/php/includes;C:/phpdev/php/class') in c:\phpdev\www\layout.php on line 44
I should note I'm using a pre-setup package called phpdev... BUT, even if I do a full manual install of mysql/php/apache I still get a similar error with different paths...
ssoooooo.... Does any guru out there have an answer? I've wracked my brains, I've searched high and low, I just am about to give up (but can't, seeing as this PHP stuff is so liberating). I'd really appreciate any (ANY) ideas
Kindest regards,
Rob
I've installed many times php, mysql and apache (1.3x and 2.x) on my win2k home machine so I can do my dev testing at home before screwin up my server
On my original installation I managed to get it all down well with a foxserv installation but something went wrong, I uninstalled, re-installed (many times) and it kept erroring..
Here is the problem:
I have been using PHP to create a tables based page that has a menu on the left (which is included into a layout) and the content on the right (which is included into the same layout).
The way I do it is to pass a variable from my menu to the layout page which then places the content on the right according to the variable passed.
I presume this is a common way of doing things, as I'm a newby I was greatly impressed with this simple feature alone to make managing large sites simpler...
OK, back to the problem...
Now I've re-installed the software (apache/mysql/php) it all works fine except the ability to pass the variable and accept it to re-generate the page with the appropriate content.
Here is the code I use in my menu:
Code: Select all
<a href="layout.php?id=content.php">Link name</a>I haven't included <? and ?> in any of these examples, please imagine they exist
Code: Select all
$interview = "$id";
include("$interview");NOW THOUGH.. this is the error I get:
Warning: Failed opening '' for inclusion (include_path='.;C:/phpdev/php/includes;C:/phpdev/php/class') in c:\phpdev\www\layout.php on line 44
I should note I'm using a pre-setup package called phpdev... BUT, even if I do a full manual install of mysql/php/apache I still get a similar error with different paths...
ssoooooo.... Does any guru out there have an answer? I've wracked my brains, I've searched high and low, I just am about to give up (but can't, seeing as this PHP stuff is so liberating). I'd really appreciate any (ANY) ideas
Kindest regards,
Rob