It will probably behave the same way include does, which is correct
behavior.

$file=include(whatever) is telling php to store the result from include in
the $file variable, not storing a command in it.

You could do $file='include()', and then you would wind up with a variable
with whatever text you assigned to it.

This is covered in the manual,

Jason

-----Original Message-----
From: Sascha Braun [mailto:[EMAIL PROTECTED]] 
Sent: Thursday, August 15, 2002 10:31 AM
To: PHP Mailingliste; Jay Blanchard
Subject: Re: [PHP] include opens source, but it shouldn't

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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