As a matter of code length could you just append it to $information ? $information .= Printf("%c%c",63,62); instead of $information = $information . Printf("%c%c",63,62); Patrick Sent from my Verizon Wireless BlackBerry -----Original Message----- From: "Atkinson, Robert" <ratkinson@xxxxxxxxxxxxx> Date: Thu, 24 Apr 2008 08:55:11 To:<php-objects@xxxxxxxxxxxxxxx> Subject: RE: Include file OK, you're not writing the PHP tags to the inc file. Change you line to :- $information = $information . printf($format,$first,$second); You will also need to close the tags with :- $information = $information . Printf("%c%c",63,62); Rob. -----Original Message----- From: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com [mailto:php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com] On Behalf Of Michael Reaves Sent: 23 April 2008 21:02 To: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com Subject: RE: Include file Robert, This is what the file looks like $host='localhost'; $user='root'; $pwd=''; $dv='dds'; Michael J. Reaves, MS, MCP Project Specialist _____ From: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com [mailto:php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com] On Behalf Of Atkinson, Robert Sent: Wednesday, April 23, 2008 2:23 PM To: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com Subject: RE: Include file Michael - I can't see any obvious problems with your code. What does data.inc look like - does it contain the right information? Robert. ________________________________ From: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com on behalf of Michael Reaves Sent: Wed 23/04/2008 20:01 To: php-objects@ <mailto:php-objects%40yahoogroups.com> yahoogroups.com Subject: Include file Group, I am trying to create an include file that contains the info to connect to a MySQL database. I need the file look like this; <?php $host = 'localhost'; $user = 'root'; $pwd = ''; $db = 'dvd'; ?> Here is my code so far; <?php $file = "data.inc"; $format = "%c%cphp"; $first=60; $second=63; printf($format,$first,$second); $information = $information."\$host='$server';\n"; $information = $information."\$user='$user';\n"; $information = $information."\$pwd='$pwd';\n"; $information = $information."\$dv='$db';\n"; echo $information; $fw = fopen($file,"w+") or die("Can't open file $file"); $fout = fwrite($fw,$information); fclose($fw); ?> I have tried char and printf. When I run this the <?php must be interpreted as a new php file and nothing is output. Any suggestions will be appreciated. Michael J. Reaves, MS, MCP Project Specialist [Non-text portions of this message have been removed] ------------------------------------ Are you looking for a PHP job? Join the PHP Professionals directory Now! http://www.phpclass <http://www.phpclass> <http://www.phpclass <http://www.phpclasses.org/professionals/Yahoo> es.org/professionals/Yahoo> es.org/professionals/Yahoo! Groups Links **************************************************************************** ******* Any opinions expressed in email are those of the individual and not necessarily those of the company. This email and any files transmitted with it are confidential and solely for the use of the intended recipient or entity to whom they are addressed. It may contain material protected by attorney-client privilege. 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