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Posted by ZeldorBlat on 05/17/07 18:22
On May 17, 1:38 pm, Justin Voelker <justin.voel...@gmail.com> wrote:
> I have a configuration file that contains the host, username,
> password, and database name required for any database connections.
> The config file runs through a few if/then/else statements to
> determine if the website is local or remote and if it is local if it
> is the test site or "live" site then it sets the 4 variables. I am
> trying to implement a webstats package that has it's own config file.
> I would the new config file to be able to use the same variables as my
> config file. The problem I am encountering is that the new config
> file has all of its variables inside of a class and I can't seem to
> get to my variables from within it. Right now the file starts like
> this:
>
> <?php
> class SlimStatConfig {
> /** Database connection */
> var $server = "localhost";
> var $username = "username";
> var $password = "password";
> var $database = "database";
> ...
> ?>
>
> What I am envisioning is something to the effect of:
>
> <?php
> include('/additionalfiles/config.php');
> class SlimStatConfig {
> /** Database connection */
> global $dbhost;
> global $dbuser;
> global $dbpasswd;
> global $dbname;
> var $server = $dbhost;
> var $username = $dbuser;
> var $password = $dbpasswd;
> var $database = $dbname;
> ...
> ?>
>
> I have never worked with classes before. And please spare me the "how
> do you call yourself a php developer without knowing how to use
> classes." Right now I just would like to get this to work with a
> minimal amount of work. Thanks for your help!
Does the class have a constructor? If so, that's the place to pass in
the variables. It might look something like this:
public function __construct($dbhost, $dbuser, $dbpasswd, $dbname) {
$this->dbhost = $dbhost;
$this->dbuser = $dbuser;
$this->dbpasswd = $dbpasswd;
$this->dbname = $dbname;
}
Then, when you instantiate the class, you would do it like this:
$s = new SlimStatConfig($dbhost, $dbuser, $dbpasswd, $dbname);
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