Can't you use a fopen('$file','r+'); or some like this instead?
Otherwise, I don't use vars to store includes in.
I always use constructs like this:
include('/path-to-file/file-name.inc.php');
But what happens when you put a file via fopen() driektive into an array? -
Will it be parsed later on?
Schura
----- Original Message -----
From: "Jay Blanchard" <[EMAIL PROTECTED]>
To: "'Harry.de'" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, August 15, 2002 6:25 PM
Subject: RE: [PHP] include opens source, but it shouldn't
> [snip]
> why does include always opens my included file
> although it is defined as a variable
>
> Example:
>
> $file=include(/myfile/myfile.txt/);
>
> shouldn't it only be opend by typing
>
> echo $file;
> [/snip]
>
> The include() opens the file to place it in the variable, in other words
> include means READ INTO. Therfore
> $file=READ INTO THIS FILE(/myfile/myfile.txt/);
>
> HTH!
>
> Jay
>
> I'm really easy to get along with, once you people learn to worship me
>
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