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@xxxxxxxxxxxxxxx [mailto:php-objects@xxxxxxxxxxxxxxx] On
Behalf Of Michael Reaves
Sent: 23 April 2008 21:02
To: php-objects@xxxxxxxxxxxxxxx
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@xxxxxxxxxxxxxxx [mailto:php-objects@xxxxxxxxxxxxxxx] On
Behalf Of Atkinson, Robert
Sent: Wednesday, April 23, 2008 2:23 PM
To: php-objects@xxxxxxxxxxxxxxx
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
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