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:
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<a href="layout.php?id=content.php">Link name</a>I haven't included <? and ?> in any of these examples, please imagine they exist
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$interview = "$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