New Php Problem

[Deno]

Hello,

 

I have another problem now:

 

This is relating back to my interviews.php page. When I try to select a link, it won't print the page contents. In my database "interviews", I have the following fields: interview_id, name and page. Page contains the filename of the page, that page mainly has one image and a bunch of text.

 

Is there a way PHP can print php files? Any help would be appreciated, thanks.

[borfast]

Perhaps you can include the whole file?

It's the simplest trick of all, just place this: include("page"); where you want the script to "print" the file and it will be there.

 

If you want it to print the page on another location, that's a different story.

[DarqFlare]

I'm with Raul here. Back when I used a ?Page=filename.php system for managing pages on my websites, I would just use require($Page); to load it up.

 

Something else to do is fopen($Page); then print its contents.

[borfast]

Deno, I forgot about one thing:

 

be carefull when using such a method for including pages. Someone might try to retrieve files they shouldn't be able to access simply by putting the filename in the URL.

 

Use a numbering method instead. like index.php?page=x, where x is a number and then, on index.php you'd have:

>switch($_GET['page'])
{
  case 1:
  include "page1.php";
  break;

  case 2:
  include "page2.php"
  break;

  ...

  default:
  // Do nothing or include a default page here
}

 

Hope this helps

Edited December 5, 2003 by TCH-Raul

[borfast]

Oh, one more thing: don't use just $Page to get the variable passed in the URL. If 'register_globals' is turned off, your script will fail. Instead, use $_GET['Page']. You should always use $_GET['variable_name'] for variables passed in the URL, $_POST['variable_name'] for variables sent through an HTML form, $_COOKIES['cookie_name'] for cookies, etc... because this way you know your script will always work, whether register_globals is on or off.

Edited December 5, 2003 by TCH-Raul