Re: 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" <jay.blanchard@niicommunications.com> To: "'Harry.de'" <harald.mohring@gmx.de>; <php-general@lists.php.net> 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 > > *********************************************************** > * Texas PHP Developers Conf Spring 2003 * > * T Bar M Resort & Conference Center * > * New Braunfels, Texas * > * San Antonio Area PHP Developers Group * > * Interested? Contact jay.blanchard@niicommunications.com * > *********************************************************** > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >