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put contents of linked file into section of document

Posted by Christina on 05/07/06 23:52

I'm pretty new to using PHP, so please bear with me.

This subject has probably been covered a hundred times, but I can't find a
clear answer on the web or in a newsgroup. I have a 650 page intranet site
that I am working on with categories and subcategories (each in their own
subsequent folders) and I am using PHP (4.1+) and MySQL for populating
links. There will be a dynamically populated menu from the database. The
users will be uploading files whose links will be added to the database
through a form.

I have the header, navigation, and footer sections in their own PHP/HTML
files to be included on page with the <?PHP include
("includes/filename.html"); ?>. This part is easy, but I have to create a
whole new page for each and every web page that includes this code. What I
need to do is have the basic .html page show up in the content section of
the current page when a person clicks a link in the navigation menu or
current page. Essentially I would want it to work like a framed page (<a
href: "linkedpage.html" target="content">, but I don't want to use frames or
iframes.

I would also like to have the page that was inserted be reflected as a URL
in the browser location for bookmarking, as well as have the page title be
automatically assigned. In my search on the web I find a lot of reference to
combining PHP and Javascript. I believe I've seen it done with Javascript
using the GetElementByID method, but it's been a while since I've written
Javascript.

I'm sure there is a simple basic way to format each link itself in just PHP.
Should I assign a variable to hold the value of the link? I assume it
includes having a <div id="content">. Should I format the link or the
content section with HTTP_REFERRER or use File_Include_Path or does this
have to do with fopen_wrapper? There are so many different things I've
seen, but nothing conclusive, that my head is spinning.

Thanks,

Chris

 

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